[leetcode] 19. Remove Nth Node From End of List

转载于 2016-07-28 06:43:31 发布 · 164 阅读

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本文介绍了一种高效算法，用于在一次遍历中删除给定链表的倒数第N个节点，并提供了详细的实现步骤及C++代码示例。

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.

解法一：

让cur先走n步，如果走到null，则说明需要移除head。不然，再让cur和pre一起移动，当cur移动到null时，pre所在的node就是倒数第n个node。然后简单的操作pre->next即可。

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