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[Linked List]Remove Nth Node From End of List

时间:2015-12-12 11:00:36      阅读:119      评论:0      收藏:0      [点我收藏+]
Total Accepted: 84303 Total Submissions: 302714 Difficulty: Easy

 

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

For 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.
Try to do this in one pass.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
 /*
 求长度
 找前驱
 删除
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        int len = 0;
        ListNode* p = head;
        while(p){
            len++;
            p=p->next;
        }
        p = head;
        ListNode* pre = NULL;
        int m = len -n;
        while(m){
            pre = p;
            p=p->next;
            m--;
        }
        if(pre){
            pre->next = p->next;
        }else{
            p = head;
            head = head->next;
        }
        delete(p);
        return head;
    }
};

[Linked List]Remove Nth Node From End of List

原文:http://www.cnblogs.com/zengzy/p/5040789.html

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