題目

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.

分析

因?yàn)槿ブ氐氖且粋€(gè)已經(jīng)排好序的數(shù)組，所以有重復(fù)的數(shù)據(jù)一定在一起，遍歷數(shù)組，判斷當(dāng)前數(shù)據(jù)是否有重復(fù)，有重復(fù)將重復(fù)的數(shù)據(jù)用后面的數(shù)據(jù)覆蓋

代碼

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