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【LeetCode】19. Remove Nth Node From End of List

编程语言 zjajgyy 11℃ 0评论
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1.题目

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

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.

2.思路

先从头遍历n个结点,找到需要删除的前一个结点removePre,然后每次移动一次游标,移动一次removePre。注意如果删除的是head,需要做处理,因为removePre此时应该没有值。

3.代码

/**
 * 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* next = head;
        ListNode* removePre = NULL;
        int firstN = n;
        while (n > 0) {
            next = next->next;
            n--;
        }
        if (next!=NULL) {
            removePre = head;
        } else {
            return head->next;
        }

        while (next->next != NULL) {
            next = next->next;
            removePre = removePre->next;
        }
        ListNode* tmp = removePre->next;
        removePre->next = (tmp==NULL)?tmp:tmp->next;
        return head;
    }
};

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