LC 92. 反转链表 II

题目描述

给你单链表的头节点 head 和两个整数 left 和 right ，其中 left <= right 。请你反转从位置 left 到位置 right 的链表节点，返回 反转后的链表 。

样例

输入：head = [1,2,3,4,5], left = 2, right = 4
输出：[1,4,3,2,5]

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做链表题目的关键是 画图、画图、画图！

解决链表翻转类问题的思路：需要3个pointer辅助

1.数组记录并交换存储的值

Java 代码

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {

    ListNode[] list = new ListNode [505];
    int idx = 1;

    public ListNode reverseBetween(ListNode head, int left, int right) {
        ListNode h = head;
        int mid = (left + right) / 2;
        if ((left + right) % 2 == 0) {
            for (int i = 1; i <= right; i ++ , h = h.next) {
                if (i >= left && i < mid) {
                    list[idx ++ ] = h;
                }
                else if (i > mid) {
                    int x = idx - (i - mid);
                    int temp = list[x].val;
                    list[x].val = h.val;
                    h.val = temp;
                }
            }
        } else {
            for (int i = 1; i <= right; i ++ , h = h.next) {
                if (i >= left && i <= mid) {
                    list[idx ++ ] = h;
                }
                else if (i > mid) {
                    int x = idx - (i - mid);
                    int temp = list[x].val;
                    list[x].val = h.val;
                    h.val = temp;
                }
            }
        }
        return head;
    }
}

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2.节点指针反向操作

Java 代码

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {


    public ListNode reverseBetween(ListNode head, int left, int right) {
        ListNode dummy = new ListNode(-1);
        dummy.next = head;

        ListNode a = dummy;
        for (int i = 0; i < left - 1; i ++ ) a = a.next; // left的前一个节点

        ListNode b = a.next; // left
        ListNode c = b.next;

        for (int i = 0; i < right - left; i ++ ) {
            ListNode t = c.next;

            c.next = b;
            b = c;
            c = t;
        }

        a.next.next = c;
        a.next = b;

        return dummy.next;

    }
}