题目

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

这么一个简单的题目，竟然3次才AC。太失败了！

总结，就是只顾得求出不重复元素的个数，却忽略了输入参数是引用，必须同时使得数组中存储为不重复元素。也就是求取数目的同时必须更新数组中元素不能重复。

AC代码

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if (nums.size() <= 1)
            return nums.size();

        int i = 0, j = 1;
        while (j < nums.size())
        {
            if (nums[i] == nums[j])
            {
                ++j;
            }
            else
            {
                nums[++i] = nums[j++];
            }
        }
        return i + 1;
    }
};

GitHub测试程序源码