【LeetCode】Reverse Linked List II 解题报告

标签：链表   反转   reverse linked list   

【题目】

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

题意：把链表中m到n的部分反转（1<=m<=n<=length）。

注意要求：在原地反转，也就是不能申请额外的空间，且只能遍历一遍。

自然而然想到把链表分为三部分，重点是如何遍历一遍把中间部分反转？借助两个指针，tmpHead和tmpNext，tmpHead是遍历反转后的head，tmpNext始终是tmpHead反转后的next。

直接看代码吧，代码中有注释，不懂的请留言。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    // Divide the list to three parts: 
    // 1)Nodes before m keep still; 
    // 2)Traverse m~n nodes;
    // 3)Nodes after n keep still.
    public ListNode reverseBetween(ListNode head, int m, int n) {
        if (m == n) return head;
        
        ListNode preHead = new ListNode(0);
        preHead.next = head;
        
        // The (m-1) node is the tail of first tail.
        ListNode firstTail = preHead;
        int k = m - 1;
        while (k-- > 0) {
            firstTail = firstTail.next;
        }
        
        // The m-th node is the traversed tail.
        ListNode secondTail = firstTail.next;
        
        // Traverse m~n nodes, and get the traversed head.
        ListNode tmpHead = null;
        ListNode tmpNext = null;
        ListNode node = firstTail.next;
        k = n - m + 1;
        while (k-- > 0) {
            tmpHead = node;
            node = node.next;
            
            tmpHead.next = tmpNext;
            tmpNext = tmpHead;
        }
        
        // Connect three parts.
        firstTail.next = tmpHead;
        secondTail.next = node;
        
        return preHead.next;
    }
}

指针关系比较复杂，想清楚了在写代码。

【LeetCode】Reverse Linked List II 解题报告

标签：链表   反转   reverse linked list   

原文地址：http://blog.csdn.net/ljiabin/article/details/41748689