Remove Duplicates from Sorted Array

最新推荐文章于 2022-04-06 16:22:53 发布

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本文介绍了一个C++程序，该程序可以就地移除已排序数组中的重复元素，并返回新数组的有效长度。通过遍历数组并仅保留不同元素的方法，实现了常数空间复杂度的要求。

Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.

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

For example,
Given input array nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn't matter what you leave beyond the new length.

#include<iostream>
#include<vector>

using namespace std;

int removeDuplicates(vector<int>& nums) {
int num = 0;//用于计算所有重复的数的个数
int n = nums.size();
for(int a = 1; a < n; a++)
{
if(nums[a] == nums[a - 1]) num++;//每有一个重复的数，num++
else nums[a - num] = nums[a]; //如果和前面的数不重复，则将这个数放到新的数组的下一位
}
return n - num;//新的数组的个数大小即为原来数组的大小减去重复的数的个数
}

int main()
{
int n;
vector<int> q;
for(int a = 0;a < 5;a++){
cin >> n;
q.push_back(n);
}
int w = removeDuplicates(q);
for(int a = 0; a<5;a++) cout << q[a] <<" ";
cout << w;
return 0;
}