【Leetcode】Remove Nth Node From End of List

题目描述：

Given a linked list, remove the n-th node from the end of list and return its head.

Example:

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:

Given n will always be valid.

Follow up:

Could you do this in one pass?

理解：

　　定义两个指针，一个在前，一个在后，同时遍历ListNode，要求是：开始遍历时fast指针要比slow指针快n步，方法是先让fast指针先移动n步，然后fast和slow再同时移动，直到移动到末尾。

这里有两种解决思路，一种直接在ListNode上进行操作，另一种使用一个stack。

解法一：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if(!head) return NULL;
        ListNode* new_head = new ListNode(0);
        new_head->next = head;
        ListNode* slow = new_head, *fast = new_head;
        for(int i = 0;i<n;++i){  //fast先移动n
            fast = fast->next;
        }
        if(!fast) return new_head->next;
        while(fast->next){      //一起移动
            fast = fast->next;
            slow = slow->next;
        }
        slow->next = slow->next->next;
        return new_head->next;
    }
};

解法二：利用stack来解决

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if(!head) return NULL;
        stack<ListNode*> s;
        ListNode* node = head;
       while(node){
           s.push(node);
           node = node->next;
       }
        for(int i = 0; i < n; ++i) {
            s.pop();
        }
       if (s.empty()) {
            return head->next;
        } else {
            s.top()->next = s.top()->next->next;
            return head;
        }
    }
};

完