leetcode 26. Remove Duplicates from Sorted Array

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本文介绍了一种在不使用额外空间的情况下去除有序数组中重复元素的方法，并保持每个元素只出现一次。通过遍历数组并将不重复的元素放置在数组前端来实现。详细介绍了算法的具体实现过程。

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

解：由于是已经排好序的，所以相同的数字只会按照顺序出现，遍历一遍，将不同的数字放在数组的前面并记录长度即可。

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