链表－leetcode 92 Reverse Linked List II

原题链接：Reverse Linked List II

题意比较明晰，代码逻辑也比较清晰，思路就是，标记第m个节点，同时将m到n的节点的值入栈，进行反转节点

代码如下

/**
 * 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) {
        /*
            思路：标记坐标m指向的节点，借助栈将m到n之间的节点反转
            Time Complexity:O(n)
            Space Complexity:O(m-n)
        */
        if(!head || !head->next)return head;
        ListNode* ret=head;
        ListNode* start=NULL;
        stack<int>sta;
        int i=1;
        while(i<=n){
            if(i<m){
                head=head->next;
                i++;
                continue;
            }
            if(i==m){
                start=head;
                sta.push(head->val);
                head=head->next;
                i++;
                continue;
            }
            if(i<n){
                sta.push(head->val);
                head=head->next;
                i++;
                continue;
            }
            if(i==n){
                sta.push(head->val);
                break;
            }
        }
        while(!sta.empty()){
            start->val=sta.top();
            sta.pop();
            start=start->next;
        }
        return ret;
    }
};