leetcode_19_Remove Nth Node From End of List (easy)

标签：

Remove Nth Node From End of List

题目：

Given a linked list, remove the n^th 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.

解体：
应该再一次循环，但是不知道链表总共多少数据

简单的两次循环，但如何才能预知未来>_<

（提示：英文翻译错误，哎，我的英文是体育老师教的）

#include <iostream>
using namespace std;

struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
 };

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode *now = head;
        int i = 0;
        for(;now!=NULL;now = now->next){
            i++;
        }
        if(n>i){
            return NULL;
        }
        if(n==i){
            return head->next;
        }
        int j = 0;
        for(now=head;now!=NULL;now = now->next){
            j++;
            if(j == i-n){
                ListNode *next = now->next->next;
                delete now->next;
                now->next = next;
                break;
            }
        }
        return head;
    }
};

void printList(ListNode *head){
    ListNode *now = head;
    for(;now!=NULL;now = now->next){
         cout<<now->val<<"->";
    }
}

int main(int argc, const char * argv[]) {
    ListNode *head = (ListNode *)malloc(sizeof(ListNode));
    head->val = 1;
    head->next = NULL;
    ListNode *now = head;
    for(int i=2;i<6;i++){
        ListNode *next = (ListNode *)malloc(sizeof(ListNode));
        next->val = i;
        next->next = NULL;
        now->next = next;
        now = next;
    }
    cout<<"before"<<endl;
    printList(head);
    cout<<endl;
    Solution a;
    ListNode *final = a.removeNthFromEnd(head, 2);
    cout<<"after"<<endl;
    printList(final);
    return 0;
}
什么，尽然过了，what

leetcode_19_Remove Nth Node From End of List (easy)

标签：

原文地址：http://www.cnblogs.com/ganeveryday/p/4903452.html