06链表问题

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1 链表问题

面试时链表解题的方法论

对于笔试，不用太在乎空间复杂度，一切为了时间复杂度

对于面试，时间复杂度依然放在第一位，但是一定要找到空间最省的方法

1.1 链表面试常用数据结构和技巧

1、 使用容器（哈希表，数组等）

2、 快慢指针

1.1.1 快慢指针问题

1、 输入链表头结点，奇数长度返回中点，偶数长度返回上中点

1 3 5 2 7 返回 5；1 3 2 7 返回 3

2、 输入链表头结点，奇数长度返回中点，偶数长度返回中下点

1 3 5 2 7 返回 5；1 3 2 7 返回 2

3、 输入链表头结点，奇数长度返回中点前一个，偶数长度返回上中点前一个

1 3 5 2 7 返回 3；1 3 2 7 返回 1

4、 输入链表头结点，奇数长度返回中点前一个，偶数长度返回下中点前一个

1 3 5 2 7 返回 3；1 3 2 7 返回 3

package class06;

import java.util.ArrayList;

public class Code01_LinkedListMid {

	public static class Node {
		public int value;
		public Node next;

		public Node(int v) {
			value = v;
		}
	}

	// head 头
	// 1、奇数长度返回中点，偶数长度返回上中点
	public static Node midOrUpMidNode(Node head) {
	  // 没有点，有一个点，有两个点的时候都是返回头结点
		if (head == null || head.next == null || head.next.next == null) {
			return head;
		}
		// 链表有3个点或以上
		// 快慢指针，快指针一次走两步，慢指针一次走一步
		// 快指针走完，慢指针在中点位置
		Node slow = head.next;
		Node fast = head.next.next;
		while (fast.next != null && fast.next.next != null) {
			slow = slow.next;
			fast = fast.next.next;
		}
		return slow;
	}

        // 2、奇数长度返回中点，偶数长度返回中下点
	public static Node midOrDownMidNode(Node head) {
		if (head == null || head.next == null) {
			return head;
		}
		Node slow = head.next;
		Node fast = head.next;
		while (fast.next != null && fast.next.next != null) {
			slow = slow.next;
			fast = fast.next.next;
		}
		return slow;
	}

        // 3、奇数长度返回中点前一个，偶数长度返回上中点前一个
	public static Node midOrUpMidPreNode(Node head) {
		if (head == null || head.next == null || head.next.next == null) {
			return null;
		}
		Node slow = head;
		Node fast = head.next.next;
		while (fast.next != null && fast.next.next != null) {
			slow = slow.next;
			fast = fast.next.next;
		}
		return slow;
	}

        // 4、奇数长度返回中点前一个，偶数长度返回下中点前一个
	public static Node midOrDownMidPreNode(Node head) {
		if (head == null || head.next == null) {
			return null;
		}
		if (head.next.next == null) {
			return head;
		}
		Node slow = head;
		Node fast = head.next;
		while (fast.next != null && fast.next.next != null) {
			slow = slow.next;
			fast = fast.next.next;
		}
		return slow;
	}

        // 笔试可以用这种复杂点的方法，空间复杂度比上面快慢指针要高
	public static Node right1(Node head) {
		if (head == null) {
			return null;
		}
		Node cur = head;
		ArrayList<Node> arr = new ArrayList<>();
		while (cur != null) {
			arr.add(cur);
			cur = cur.next;
		}
		return arr.get((arr.size() - 1) / 2);
	}

	public static Node right2(Node head) {
		if (head == null) {
			return null;
		}
		Node cur = head;
		ArrayList<Node> arr = new ArrayList<>();
		while (cur != null) {
			arr.add(cur);
			cur = cur.next;
		}
		return arr.get(arr.size() / 2);
	}

	public static Node right3(Node head) {
		if (head == null || head.next == null || head.next.next == null) {
			return null;
		}
		Node cur = head;
		ArrayList<Node> arr = new ArrayList<>();
		while (cur != null) {
			arr.add(cur);
			cur = cur.next;
		}
		return arr.get((arr.size() - 3) / 2);
	}

	public static Node right4(Node head) {
		if (head == null || head.next == null) {
			return null;
		}
		Node cur = head;
		ArrayList<Node> arr = new ArrayList<>();
		while (cur != null) {
			arr.add(cur);
			cur = cur.next;
		}
		return arr.get((arr.size() - 2) / 2);
	}

	public static void main(String[] args) {
		Node test = null;
		test = new Node(0);
		test.next = new Node(1);
		test.next.next = new Node(2);
		test.next.next.next = new Node(3);
		test.next.next.next.next = new Node(4);
		test.next.next.next.next.next = new Node(5);
		test.next.next.next.next.next.next = new Node(6);
		test.next.next.next.next.next.next.next = new Node(7);
		test.next.next.next.next.next.next.next.next = new Node(8);

		Node ans1 = null;
		Node ans2 = null;

		ans1 = midOrUpMidNode(test);
		ans2 = right1(test);
		System.out.println(ans1 != null ? ans1.value : "无");
		System.out.println(ans2 != null ? ans2.value : "无");

		ans1 = midOrDownMidNode(test);
		ans2 = right2(test);
		System.out.println(ans1 != null ? ans1.value : "无");
		System.out.println(ans2 != null ? ans2.value : "无");

		ans1 = midOrUpMidPreNode(test);
		ans2 = right3(test);
		System.out.println(ans1 != null ? ans1.value : "无");
		System.out.println(ans2 != null ? ans2.value : "无");

		ans1 = midOrDownMidPreNode(test);
		ans2 = right4(test);
		System.out.println(ans1 != null ? ans1.value : "无");
		System.out.println(ans2 != null ? ans2.value : "无");

	}

}

1.1.2 面试题一：判断回文结构

给定一个单链表的头结点head，请判断该链表是否为回文结构。回文就是正着输出和反着输出结果一样

  1. 栈的方法特别简单（笔试用）

笔试思路，以此把该链表放入栈中。再遍历该链表和栈中弹出的数比对，只要有不一样，就不是回文

  1. 改原链表的方法需要注意边界问题（面试用）

快慢指针解法：用快慢指针定位到中点的位置，奇数就是定位到唯一的中点，偶数定位到上中点。然后把中点右半部分加入栈中去，那么栈中存的是右半部分的逆序。接着从头遍历链表，栈中有多少个元素，我们就比较多少步，如果有对不上就不是回文

快慢指针最优解，不使用容器结构(stack)，O(1)：同样的找到中点位置，把右半部分指针回指到中点。接着指针1从L位置，指针2从R位置，往中间遍历。，每步比对，如果有不一样，则不是回文。返回答案之前，把中点右边的指针调整回来

package class06;

import java.util.Stack;

public class Code02_IsPalindromeList {

	public static class Node {
		public int value;
		public Node next;

		public Node(int data) {
			this.value = data;
		}
	}

	// need n extra space
	public static boolean isPalindrome1(Node head) {
	  // 依次进栈
		Stack<Node> stack = new Stack<Node>();
		Node cur = head;
		while (cur != null) {
			stack.push(cur);
			cur = cur.next;
		}
		// 每个元素和栈中比较
		while (head != null) {
			if (head.value != stack.pop().value) {
				return false;
			}
			head = head.next;
		}
		return true;
	}

	// need n/2 extra space
	// 中点右侧进栈
	public static boolean isPalindrome2(Node head) {
		if (head == null || head.next == null) {
			return true;
		}
		Node right = head.next;
		Node cur = head;
		while (cur.next != null && cur.next.next != null) {
			right = right.next;
			cur = cur.next.next;
		}
		Stack<Node> stack = new Stack<Node>();
		while (right != null) {
			stack.push(right);
			right = right.next;
		}
		while (!stack.isEmpty()) {
			if (head.value != stack.pop().value) {
				return false;
			}
			head = head.next;
		}
		return true;
	}

	// need O(1) extra space
	// 不使用容器（stack）的方法
	public static boolean isPalindrome3(Node head) {
		if (head == null || head.next == null) {
			return true;
		}
		// 慢指针
		Node n1 = head;
		// 快指针
		Node n2 = head;
		while (n2.next != null && n2.next.next != null) { // find mid node
			n1 = n1.next; // n1 -> mid
			n2 = n2.next.next; // n2 -> end
		}
		// n1 中点
		
		
		n2 = n1.next; // n2 -> right part first node
		n1.next = null; // mid.next -> null
		Node n3 = null;
		// 右半部逆序指向中点
		while (n2 != null) { // right part convert
			n3 = n2.next; // n3 -> save next node
			n2.next = n1; // next of right node convert
			n1 = n2; // n1 move
			n2 = n3; // n2 move
		}
		// 引入n3记录最后的位置，之后把右半部再逆序回原来的次序
		n3 = n1; // n3 -> save last node
		n2 = head;// n2 -> left first node
		boolean res = true;
		while (n1 != null && n2 != null) { // check palindrome
			if (n1.value != n2.value) {
				res = false;
				break;
			}
			n1 = n1.next; // left to mid
			n2 = n2.next; // right to mid
		}
		n1 = n3.next;
		n3.next = null;
		// 把右半部分再逆序回来
		while (n1 != null) { // recover list
			n2 = n1.next;
			n1.next = n3;
			n3 = n1;
			n1 = n2;
		}
		return res;
	}

	public static void printLinkedList(Node node) {
		System.out.print("Linked List: ");
		while (node != null) {
			System.out.print(node.value + " ");
			node = node.next;
		}
		System.out.println();
	}

	public static void main(String[] args) {

		Node head = null;
		printLinkedList(head);
		System.out.print(isPalindrome1(head) + " | ");
		System.out.print(isPalindrome2(head) + " | ");
		System.out.println(isPalindrome3(head) + " | ");
		printLinkedList(head);
		System.out.println("=========================");

		head = new Node(1);
		printLinkedList(head);
		System.out.print(isPalindrome1(head) + " | ");
		System.out.print(isPalindrome2(head) + " | ");
		System.out.println(isPalindrome3(head) + " | ");
		printLinkedList(head);
		System.out.println("=========================");

		head = new Node(1);
		head.next = new Node(2);
		printLinkedList(head);
		System.out.print(isPalindrome1(head) + " | ");
		System.out.print(isPalindrome2(head) + " | ");
		System.out.println(isPalindrome3(head) + " | ");
		printLinkedList(head);
		System.out.println("=========================");

		head = new Node(1);
		head.next = new Node(1);
		printLinkedList(head);
		System.out.print(isPalindrome1(head) + " | ");
		System.out.print(isPalindrome2(head) + " | ");
		System.out.println(isPalindrome3(head) + " | ");
		printLinkedList(head);
		System.out.println("=========================");

		head = new Node(1);
		head.next = new Node(2);
		head.next.next = new Node(3);
		printLinkedList(head);
		System.out.print(isPalindrome1(head) + " | ");
		System.out.print(isPalindrome2(head) + " | ");
		System.out.println(isPalindrome3(head) + " | ");
		printLinkedList(head);
		System.out.println("=========================");

		head = new Node(1);
		head.next = new Node(2);
		head.next.next = new Node(1);
		printLinkedList(head);
		System.out.print(isPalindrome1(head) + " | ");
		System.out.print(isPalindrome2(head) + " | ");
		System.out.println(isPalindrome3(head) + " | ");
		printLinkedList(head);
		System.out.println("=========================");

		head = new Node(1);
		head.next = new Node(2);
		head.next.next = new Node(3);
		head.next.next.next = new Node(1);
		printLinkedList(head);
		System.out.print(isPalindrome1(head) + " | ");
		System.out.print(isPalindrome2(head) + " | ");
		System.out.println(isPalindrome3(head) + " | ");
		printLinkedList(head);
		System.out.println("=========================");

		head = new Node(1);
		head.next = new Node(2);
		head.next.next = new Node(2);
		head.next.next.next = new Node(1);
		printLinkedList(head);
		System.out.print(isPalindrome1(head) + " | ");
		System.out.print(isPalindrome2(head) + " | ");
		System.out.println(isPalindrome3(head) + " | ");
		printLinkedList(head);
		System.out.println("=========================");

		head = new Node(1);
		head.next = new Node(2);
		head.next.next = new Node(3);
		head.next.next.next = new Node(2);
		head.next.next.next.next = new Node(1);
		printLinkedList(head);
		System.out.print(isPalindrome1(head) + " | ");
		System.out.print(isPalindrome2(head) + " | ");
		System.out.println(isPalindrome3(head) + " | ");
		printLinkedList(head);
		System.out.println("=========================");

	}

}

1.1.3 面试题二：按值划分单链表

将单链表按某值划分成左边小，中间相等，右边大的形式

  1. 把链表放入数组里，在数组上做partion(笔试用)

[3, 2, 4, 7, 0, 2, 1]选择2划分，基于2对数组作partion

  1. 分成小、中、大三部分。再把各个部分之间串起来（面试用）

借助6个引用变量，不需要容器O(N)，且能保证稳定性。小于区域的头引用，小于区域的尾引用，等于区域的头引用，等于区域的尾引用，大于区域的头引用，大于区域的尾引用。依次对比给定的值加入到这三个区域，之后串联起来

package class06;

public class Code03_SmallerEqualBigger {

	public static class Node {
		public int value;
		public Node next;

		public Node(int data) {
			this.value = data;
		}
	}

  // 方法1
	public static Node listPartition1(Node head, int pivot) {
		if (head == null) {
			return head;
		}
		Node cur = head;
		int i = 0;
		while (cur != null) {
			i++;
			cur = cur.next;
		}
		Node[] nodeArr = new Node[i];
		i = 0;
		cur = head;
		for (i = 0; i != nodeArr.length; i++) {
			nodeArr[i] = cur;
			cur = cur.next;
		}
		arrPartition(nodeArr, pivot);
		for (i = 1; i != nodeArr.length; i++) {
			nodeArr[i - 1].next = nodeArr[i];
		}
		nodeArr[i - 1].next = null;
		// 返回头结点
		return nodeArr[0];
	}

	public static void arrPartition(Node[] nodeArr, int pivot) {
		int small = -1;
		int big = nodeArr.length;
		int index = 0;
		while (index != big) {
			if (nodeArr[index].value < pivot) {
				swap(nodeArr, ++small, index++);
			} else if (nodeArr[index].value == pivot) {
				index++;
			} else {
				swap(nodeArr, --big, index);
			}
		}
	}

	public static void swap(Node[] nodeArr, int a, int b) {
		Node tmp = nodeArr[a];
		nodeArr[a] = nodeArr[b];
		nodeArr[b] = tmp;
	}

  // 方法2
	public static Node listPartition2(Node head, int pivot) {
		Node sH = null; // small head
		Node sT = null; // small tail
		Node eH = null; // equal head
		Node eT = null; // equal tail
		Node mH = null; // big head
		Node mT = null; // big tail
		Node next = null; // save next node
		// every node distributed to three lists
		while (head != null) {
			next = head.next;
			head.next = null;
			if (head.value < pivot) {
			  // 小于节点为空，当前节点即做头又做尾
				if (sH == null) {
					sH = head;
					sT = head;
				// 老的尾节点指向当前节点，老的尾变成当前节点
				} else {
					sT.next = head;
					sT = head;
				}
			} else if (head.value == pivot) {
				if (eH == null) {
					eH = head;
					eT = head;
				} else {
					eT.next = head;
					eT = head;
				}
			} else {
				if (mH == null) {
					mH = head;
					mT = head;
				} else {
					mT.next = head;
					mT = head;
				}
			}
			head = next;
		}
		// 小于区域的尾巴，连等于区域的头，等于区域的尾巴连大于区域的头
		if (sT != null) { // 如果有小于区域
			sT.next = eH;
			eT = eT == null ? sT : eT; // 下一步，谁去连大于区域的头，谁就变成eT
		}
		// 上面的if，不管跑了没有，et
		// all reconnect
		if (eT != null) { // 如果小于区域和等于区域，不是都没有
			eT.next = mH;
		}
		return sH != null ? sH : (eH != null ? eH : mH);
	}

	public static void printLinkedList(Node node) {
		System.out.print("Linked List: ");
		while (node != null) {
			System.out.print(node.value + " ");
			node = node.next;
		}
		System.out.println();
	}

	public static void main(String[] args) {
		Node head1 = new Node(7);
		head1.next = new Node(9);
		head1.next.next = new Node(1);
		head1.next.next.next = new Node(8);
		head1.next.next.next.next = new Node(5);
		head1.next.next.next.next.next = new Node(2);
		head1.next.next.next.next.next.next = new Node(5);
		printLinkedList(head1);
		// head1 = listPartition1(head1, 4);
		head1 = listPartition2(head1, 5);
		printLinkedList(head1);

	}

}

1.1.4 面试题三

一种特殊的单链表结构如下：

public static class Node {
		public int value;
		public Node next;
		public Node rand;

		public Node(int data) {
			this.value = data;
		}
	}

rand指针式单链表节点结构中新增加的指针，rand可能指向链表中的任意一个节点，也可能为null。给定一个由Node节点类型组成的无环单链表节点head。请实现一个函数完成这个链表的复制，并返回复制的新链表的头结点。

要求时间复杂度为O(N)，额外空间复杂度为O(1)

  1. 哈希表方法(笔试推荐)

第一步遍历，把所有节点加入到Map<Node, Node>表示老节点到克隆出来的节点映射

第二步遍历，查map找到克隆节点，最后返回头结点

  1. 不用哈希表的方法，人为构造对应关系(面试推荐)

第一步：每个节点遍历的时候克隆出来一个新的节点加入到当前节点和其next节点的中间

第二步：此时经过第一步所有节点和其克隆节点都是串在一起的，依次拿出当前节点和其克隆节点，当前节点的rand指针指向的节点的克隆节点给当前节点克隆的节点的rand节点指向的节点。

第三步：此时老节点的rand指针没变化，克隆节点的rand指针也都指向了对应的克隆节点。此时在大的链表上分离出来原链表和克隆链表

package class06;

import java.util.HashMap;

public class Code04_CopyListWithRandom {

	public static class Node {
		public int value;
		public Node next;
		public Node rand;

		public Node(int data) {
			this.value = data;
		}
	}

  // 方法1
	public static Node copyListWithRand1(Node head) {
		HashMap<Node, Node> map = new HashMap<Node, Node>();
		Node cur = head;
		while (cur != null) {
		  // 当前节点，克隆出来一个相同值的新节点加入字典
			map.put(cur, new Node(cur.value));
			cur = cur.next;
		}
		// 当前节点从头开始
		cur = head;
		while (cur != null) {
			// cur 老
			// map.get(cur) 新
			map.get(cur).next = map.get(cur.next);
			map.get(cur).rand = map.get(cur.rand);
			cur = cur.next;
		}
		// 返回head对应的克隆节点
		return map.get(head);
	}

  // 方法二
	public static Node copyListWithRand2(Node head) {
		if (head == null) {
			return null;
		}
		Node cur = head;
		Node next = null;
		// 克隆出来的node放在原本node和next指向的node中间
		// 1 -> 2
		// 1 -> 1' -> 2
		while (cur != null) {
			// cur 老   next 老的下一个
			next = cur.next;
			cur.next = new Node(cur.value);
			cur.next.next = next;
			cur = next;
		}
		cur = head;
		Node curCopy = null;
		// set copy node rand
		// 1 -> 1' -> 2 -> 2'
		// 设置新的克隆节点间的rand节点
		while (cur != null) {
			// cur 老
			// cur.next => 新的 copy出来的节点
			next = cur.next.next;
			curCopy = cur.next;
			curCopy.rand = cur.rand != null ? cur.rand.next : null;
			cur = next;
		}
		// 老的头结点：head 新克隆出来的头结点： head.next
		Node res = head.next;
		cur = head;
		// split，分离原本节点组成的链表和克隆节点组成的链表
		while (cur != null) {
			next = cur.next.next;
			curCopy = cur.next;
			cur.next = next;
			curCopy.next = next != null ? next.next : null;
			cur = next;
		}
		return res;
	}

	public static void printRandLinkedList(Node head) {
		Node cur = head;
		System.out.print("order: ");
		while (cur != null) {
			System.out.print(cur.value + " ");
			cur = cur.next;
		}
		System.out.println();
		cur = head;
		System.out.print("rand:  ");
		while (cur != null) {
			System.out.print(cur.rand == null ? "- " : cur.rand.value + " ");
			cur = cur.next;
		}
		System.out.println();
	}

	public static void main(String[] args) {
		Node head = null;
		Node res1 = null;
		Node res2 = null;
		printRandLinkedList(head);
		res1 = copyListWithRand1(head);
		printRandLinkedList(res1);
		res2 = copyListWithRand2(head);
		printRandLinkedList(res2);
		printRandLinkedList(head);
		System.out.println("=========================");

		head = new Node(1);
		head.next = new Node(2);
		head.next.next = new Node(3);
		head.next.next.next = new Node(4);
		head.next.next.next.next = new Node(5);
		head.next.next.next.next.next = new Node(6);

		head.rand = head.next.next.next.next.next; // 1 -> 6
		head.next.rand = head.next.next.next.next.next; // 2 -> 6
		head.next.next.rand = head.next.next.next.next; // 3 -> 5
		head.next.next.next.rand = head.next.next; // 4 -> 3
		head.next.next.next.next.rand = null; // 5 -> null
		head.next.next.next.next.next.rand = head.next.next.next; // 6 -> 4

		printRandLinkedList(head);
		res1 = copyListWithRand1(head);
		printRandLinkedList(res1);
		res2 = copyListWithRand2(head);
		printRandLinkedList(res2);
		printRandLinkedList(head);
		System.out.println("=========================");

	}

}

1.1.5 面试题四

该问题和约舍夫环问题是链表问题的比较难的问题

题目描述：给定两个可能有环也可能无环的单链表，头结点head1和head2。请实现一个函数，如果两个链表相交，请返回相交的第一个节点。如果不相交，返回null

要求：如果两个链表长度之和为N，时间复杂度请达到O(N)，额为空间复杂度请达到O(1)

思路：由于是单链表，则一旦成环就结束，出不来，因为每个节点只有一个Next指针

  1. 方法一：用set把每个节点的内存地址放到set里面，如果存在相同的内存在set中存在，则就是第一个成环的节点

  2. 用快慢指针对链表遍历，那么快慢指针一定会相遇，能相遇就说明存在环。然后让慢指针停在相遇的位置，快指针回到头结点。快指针和慢指针再出发且快指针也变成一次走一步和满指针相同，再次相遇的节点就是成环节点

package class06;

public class Code05_FindFirstIntersectNode {

	public static class Node {
		public int value;
		public Node next;

		public Node(int data) {
			this.value = data;
		}
	}

	public static Node getIntersectNode(Node head1, Node head2) {
		if (head1 == null || head2 == null) {
			return null;
		}
		// head1的第一个入环节点
		Node loop1 = getLoopNode(head1);
		// head2的第一个入环节点
		Node loop2 = getLoopNode(head2);
		// 两个无环链表是否相交的情况
		// 由于每个节点只有一个next指针，则如果两个无环相交则相交之后就只剩下公共部分
		// 方法1把第一条链表放到set中，第二个链表依次查在不在该set中，第一个找到的就是
		// 方法2
		// 把链表1走到尾结点end1，记录长度l1
		// 把链表1走到尾结点end2，记录长度l2
		// 如果end1和end2的内存地址不同一定不相交
		// 如果end1==end2，则(1)长的链表从头结点先走保证和短链表相同长度的位置，再以此往下走，第一次相同节点
		// (2)则依次从尾结点出发，找第一次出现内存地址不相同的那个节点，该节点的next节点就是第一次相交的节点
		if (loop1 == null && loop2 == null) {
			return noLoop(head1, head2);
		}
		
		// 一个为空，另外一个不为空不可能相交。两个都不为空的情况下共用一个环
		if (loop1 != null && loop2 != null) {
			return bothLoop(head1, loop1, head2, loop2);
		}
		return null;
	}

	// 找到链表第一个入环节点，如果无环，返回null
	public static Node getLoopNode(Node head) {
		if (head == null || head.next == null || head.next.next == null) {
			return null;
		}
		// n1 慢  n2 快
		Node n1 = head.next; // n1 -> slow
		Node n2 = head.next.next; // n2 -> fast
		while (n1 != n2) {
			if (n2.next == null || n2.next.next == null) {
				return null;
			}
			n2 = n2.next.next;
			n1 = n1.next;
		}
		// 能相遇则跳出while，快指针回到开头，满指针停在原地
		n2 = head; // n2 -> walk again from head
		while (n1 != n2) {
		    // 此时快慢指针每次移动相同步数
			n1 = n1.next;
			n2 = n2.next;
		}
		return n1;
	}

	// 如果两个链表都无环，返回第一个相交节点，如果不想交，返回null
	public static Node noLoop(Node head1, Node head2) {
		if (head1 == null || head2 == null) {
			return null;
		}
		Node cur1 = head1;
		Node cur2 = head2;
		int n = 0;
		while (cur1.next != null) {
			n++;
			cur1 = cur1.next;
		}
		while (cur2.next != null) {
			n--;
			cur2 = cur2.next;
		}
		if (cur1 != cur2) {
			return null;
		}
		// n  :  链表1长度减去链表2长度的值
    // 谁长，谁的头变成cur1
		cur1 = n > 0 ? head1 : head2; 
    // 谁短，谁的头变成cur2
		cur2 = cur1 == head1 ? head2 : head1; 
		n = Math.abs(n);
		while (n != 0) {
			n--;
			cur1 = cur1.next;
		}
		while (cur1 != cur2) {
			cur1 = cur1.next;
			cur2 = cur2.next;
		}
		return cur1;
	}

	// 两个有环链表，返回第一个相交节点，如果不想交返回null
	// head1的入环节点是loop1，head2的入环节点是loop2
	public static Node bothLoop(Node head1, Node loop1, Node head2, Node loop2) {
		Node cur1 = null;
		Node cur2 = null;
		// 类似第一种都无环的情况
		if (loop1 == loop2) {
			cur1 = head1;
			cur2 = head2;
			int n = 0;
			while (cur1 != loop1) {
				n++;
				cur1 = cur1.next;
			}
			while (cur2 != loop2) {
				n--;
				cur2 = cur2.next;
			}
			cur1 = n > 0 ? head1 : head2;
			cur2 = cur1 == head1 ? head2 : head1;
			n = Math.abs(n);
			while (n != 0) {
				n--;
				cur1 = cur1.next;
			}
			while (cur1 != cur2) {
				cur1 = cur1.next;
				cur2 = cur2.next;
			}
			return cur1;
		} else {
		  //否则，找第一个成环节点转回自身的过程中遇到loop2,则相交，否则不相交
			cur1 = loop1.next;
			while (cur1 != loop1) {
				if (cur1 == loop2) {
					return loop1;
				}
				cur1 = cur1.next;
			}
			return null;
		}
	}

	public static void main(String[] args) {
		// 1->2->3->4->5->6->7->null
		Node head1 = new Node(1);
		head1.next = new Node(2);
		head1.next.next = new Node(3);
		head1.next.next.next = new Node(4);
		head1.next.next.next.next = new Node(5);
		head1.next.next.next.next.next = new Node(6);
		head1.next.next.next.next.next.next = new Node(7);

		// 0->9->8->6->7->null
		Node head2 = new Node(0);
		head2.next = new Node(9);
		head2.next.next = new Node(8);
		head2.next.next.next = head1.next.next.next.next.next; // 8->6
		System.out.println(getIntersectNode(head1, head2).value);

		// 1->2->3->4->5->6->7->4...
		head1 = new Node(1);
		head1.next = new Node(2);
		head1.next.next = new Node(3);
		head1.next.next.next = new Node(4);
		head1.next.next.next.next = new Node(5);
		head1.next.next.next.next.next = new Node(6);
		head1.next.next.next.next.next.next = new Node(7);
		head1.next.next.next.next.next.next = head1.next.next.next; // 7->4

		// 0->9->8->2...
		head2 = new Node(0);
		head2.next = new Node(9);
		head2.next.next = new Node(8);
		head2.next.next.next = head1.next; // 8->2
		System.out.println(getIntersectNode(head1, head2).value);

		// 0->9->8->6->4->5->6..
		head2 = new Node(0);
		head2.next = new Node(9);
		head2.next.next = new Node(8);
		head2.next.next.next = head1.next.next.next.next.next; // 8->6
		System.out.println(getIntersectNode(head1, head2).value);

	}

}

1.1.6 面试题五

题目描述：能不能不给单链表的头结点，只给想要删除的节点，就能做到在链表上把这个点删掉？

  1. 抖机灵的做法，1->2->3->4->5->null，给定3。那么根据内存地址找到3这个节点，把3下个节点赋值给自身变成4，再把自身的下一个指针指向下下个值5即可。1->2->4->5->null。缺点没把原始节点删除，只是改变了值，内存地址没被删掉而是删掉了需要删除节点的下一个内存地址。该方法无法删除链表的最后一个节点

实质上不给头结点，无法删除给定的节点。没有头结点，没法准确的连指针

    
package class06;

public class Test {

	public static class Node{
		public int value;
		public Node next;
		public Node(int v) {
			value = v;
		}
	}
	
	public static void main(String[] args) {
		Node a = new Node(1);
		Node b = new Node(2);
		Node c = new Node(3);
		
		a.next = b;
		b.next = c;
		// 实质上这里置为空没用，只是把Java栈中的变量不指向堆中的3节点
		//堆中的结构没改变，3节点并没有被删除
		c = null;
		
	}
	
}

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