[LeetCode] Remove Nth Node From End of List 快慢指针

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.

 

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  Linked List Two Pointers

 

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　　简单的快慢指针问题。

#include <iostream>
using namespace std;

/**
 * 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==NULL)  return NULL;
        ListNode * fastp = head, * slowp = head;
        for(int i =0;i<n;i++)
            fastp = fastp->next;
        if(fastp==NULL) return head->next;
        while(fastp->next!=NULL){
            fastp = fastp ->next;
            slowp = slowp ->next;
        }
        slowp->next = slowp->next->next;
        return head;
    }
};

int main()
{
    return 0;
}

 

转载于:https://www.cnblogs.com/Azhu/p/4324972.html