LeetCode27. Remove Element

最新推荐文章于 2024-02-22 13:25:37 发布

Guo_Xiaoxi

最新推荐文章于 2024-02-22 13:25:37 发布

阅读量437

点赞数

CC 4.0 BY-SA版权

分类专栏： C++ LeetCode 文章标签： leetcode c++

本文链接：https://blog.youkuaiyun.com/Guo_Xiaoxi/article/details/66968785

C++ 同时被 2 个专栏收录

81 篇文章

订阅专栏

LeetCode

38 篇文章

订阅专栏

本文介绍了一种在数组中高效移除特定值的方法，并提供了两种实现方式：一种使用双循环进行元素覆盖的方式，运行时间为3ms；另一种通过先排序再移除相同元素的方式，运行时间为6ms。文章重点在于如何利用常数空间复杂度完成操作，同时变更元素顺序以达到目的。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given an array and a value, remove all instances of that value in place and return the new length.

Do not allocate extra space for another array, you must do this in place with constant memory.

The order of elements can be changed. It doesn’t matter what you leave beyond the new length.

Example:
Given input array nums = [3,2,2,3], val = 3

Your function should return length = 2, with the first two elements of nums being 2.

Hint:

Try two pointers.
Did you use the property of “the order of elements can be changed”?
What happens when the elements to remove are rare?

1、Runtime: 3 ms

#include<vector>
#include<algorithm>
#include<iostream>
using namespace std;

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int n=nums.size();
        if(n==0){return 0;}
        int res=0;
        for(int i=0;i<n;i++){
            if(nums[i]!=val){
               nums[res] =nums[i];
               res++;
            }
        }
        return res;
    }
};

void main(){
    vector<int> v;
    v.push_back(3);
    v.push_back(2);
    v.push_back(2);
    v.push_back(3);
    Solution So;
    cout<<So.removeElement(v,3)<<endl;
}

2、Runtime: 6 ms

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int n=nums.size();
        if(n==0){return 0;}
        sort(nums.begin(),nums.end());
        int res=0;
        while(res<n){
            if(nums[res]==val){
                while(res<n&&nums[res]==val){
                    nums.erase(nums.begin()+res);
                    n--;
                }
                return n;
            }
            else{
                res++;
            }       
        }
        return n;
    }
};