【LeetCode】之Remove Duplicates from Sorted Array

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.

此题目的在于给定一个已排好序的数组，删除数组中重复的元素，只保留一个，最终返回新数组的长度，且要求不能用另外的数组进行新数组元素的存储。

思路：
1.认清边界值，当数组的长度为0时，直接返回0；
2.否则，定义新数组的初始长度length为0，每次比较length所在的下标元素和当前第i个元素是否相等；
3.若不相等，length++，并把第i个元素替换掉length所在下标的元素；
4.若相等，则不做任何操作；
5.最后返回length+1的长度即为结果，所得数组的前length+1个元素即为新数组的元素。

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