LeetCode 26. Remove Duplicates from Sorted Array

题目描述

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.
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解题思路

首先看到题目的意思要我们去重，要求就是不能开新的内存空间。也就是空间复杂度是一个常数的。
一开始我想的是用vector类的erase函数去删除，后来发现不太好。因为删除的话会让迭代器失效，后面就比较麻烦。所以我就打算用两个int型变量去干这个东西。
一个是res记录当前情况下不重复的数目
一个是i记录当前扫到的位置
遍历的过程：

if(i==0 || nums[i-1]!=nums[i]){    nums[res]=nums[i];    ++res;}

实现代码

class Solution {public:    int removeDuplicates(vector<int>& nums) {        int res=0;        for(int i=0;i<nums.size();++i){            if(i==0 || nums[i-1]!=nums[i]){                nums[res]=nums[i];                ++res;            }        }        return res;    }};