題目

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.

分析

題目給了一個(gè)鏈表的一維數(shù)組，和一個(gè)整數(shù)n，要求刪除從末尾開始數(shù)第n個(gè)節(jié)點(diǎn)，然后返回鏈表的頭。做法是，用兩個(gè)指針，一個(gè)fast指針，一個(gè)solw指針，讓slow指針永遠(yuǎn)慢fast指針n個(gè)節(jié)點(diǎn)。fast走完后，再用slow指針來刪除需要?jiǎng)h除的節(jié)點(diǎn)

代碼

/**
 * 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) {
        ListNode* h = new ListNode(0);
        h->next = head;
        ListNode* slow = h;
        ListNode* fast = h;
        while(fast->next!=NULL){
            if(n<=0){
                slow = slow->next;
            }
            fast= fast->next;
            n--;
        }
        if(slow->next!=NULL){
            slow->next = slow->next->next;
        }
        return h->next;
    }
};