[面试真题] LeetCode：Reverse Linked List II

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.

解法：需要定位待翻转部分的头与尾，以及翻转部分相邻的两个节点，同时还要考虑m和n之间的关系。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *reverseBetween(ListNode *head, int m, int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        ListNode *left = new ListNode(0);
        left->next = head;
        if(m < 1){
            return head;
        }
        int len = n-m;
        if(len < 1){
            return head;
        }
        for(int i=1; i<m; i++){
            left = left->next;
        }
        ListNode *phead = left->next;
        ListNode *pre = left->next;
        ListNode *lat = pre->next;
        ListNode *tail = lat->next;
        while(tail && len-1){
            lat->next = pre;
            pre = lat;
            lat = tail;
            tail = lat->next;
            len--;
        }
        lat->next = pre;
        phead->next = tail;
        left->next = lat;
        if(m == 1){
            return left->next;    
        }else{
            return head;
        }
    }
};

Run Status: Accepted!
Program Runtime: 20 milli secs

Progress: 44/44 test cases passed.