LeetCode:Remove Element
来源:互联网 发布:大额借贷软件 编辑:程序博客网 时间:2024/06/08 08:40
问题描述:
Given an array and a value, remove all instances of that value in place and return the new length.
The order of elements can be changed. It doesn't matter what you leave beyond the new length.
思路:
这道题开始理解错了,以为是要将数组中的元素去除,结果在上面浪费了很多时间,最后看到有人用覆盖的方法,所以就采用了,代码简单,容易理解。只要遇到和给定值不同的数,存入新的索引。
C++代码:
class Solution {public: int removeElement(vector<int>& nums, int val) { int len = 0; for(int i = 0;i < nums.size();++i){ if(nums[i] != val){ nums[len++] = nums[i]; } } return len; }};
0 0
- LeetCode:Remove Element
- LeetCode: Remove Element
- [Leetcode] Remove Element
- LeetCode: Remove Element
- leetcode 39: Remove Element
- [LeetCode] Remove Element
- Leetcode: Remove Element
- Leetcode:Remove Element
- Leetcode:Remove Element
- Leetcode:Remove Element
- [LeetCode]Remove Element
- LeetCode-Remove Element
- LeetCode - Remove Element
- LeetCode | Remove Element
- leetcode之Remove Element
- LeetCode - Remove Element
- LeetCode: Remove Element
- 【LeetCode】Remove Element
- NSXMLParser 解析代理方法
- RPG游戏(一)——环境搭建、地图显示
- ibatis批量插入对象
- Java Socket长连接异步单工保持心跳
- poj--3250--Bad Hair Day(模拟)
- LeetCode:Remove Element
- NSXMLParser 解析方法
- linpcap
- 【黑马程序员】C语言结构体
- java实现excel导出
- 【C#】利用正则表达式判断输入是否为纯数字、容器类
- POJ 2352 Stars
- html.encode('gbk','ignore')
- 卷积神经网络参数说明