leetcode 26—Remove Duplicates from Sorted Array

Given a sorted array nums, 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 by modifying the input array in-place with O(1) extra memory.

Example 1:

Given 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 returned length.

Example 2:

Given nums = [0,0,1,1,1,2,2,3,3,4],

Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.

It doesn't matter what values are set beyond the returned length.

想法：直接使用STL相关函数，先使用unique去掉重复。此时重复的元素并没有被删掉，只是移到了后面，该函数返回无重复元素的最后一个元素的地址

，然后在使用distance()函数得到个数

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {

        return distance(nums.begin(), unique(nums.begin(), nums.end()));
        
    }
};

另外手动实现版本

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;
        
    }
};