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.
解体:
注意是有序数列，如果不是有序，可能前后都有，那就不一样了
不能增加额外空间，所以在原来数组技术基础上重新赋值，并不断记录信数组位置
class Solution {
public:
int removeDuplicates(vector<int>& nums) {
if(nums.size()==0)return 0;
int length = 1;
int before = nums[0];
int now = 1;
for(int i = 1;i<nums.size();i++){
if(before!=nums[i]){
length++;
nums[now++] = nums[i];
before = nums[i];
}else{
before = nums[i];
}
}
for(int i = 0;i<nums.size();i++){
cout<<nums[i];
}
cout<<endl;
return length;
}
};
leetcode_26_ Remove Duplicates from Sorted Array (easy)
原文：http://www.cnblogs.com/ganeveryday/p/4912359.html