Array-Remove Duplicates from Sorted Array

问题描述：

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.

思路：

注意到数组是已排序的，用index指示array[0...index]为去重的数组，遍历每个元素，跳过重复的元素。

解决：

 1 class Solution {
 2 public:
 3     int removeDuplicates(vector<int>& nums) {
 4         
 5         //考虑数组为空的情况
 6         if (nums.size() == 0)
 7             return 0;
 8         
 9         int index = 0;
10 
11         //循环体中跳过重复的元素
12         for (int i = 1; i < nums.size(); i++)
13         {
14             if (nums[index] != nums[i])
15                 nums[++index] = nums[i];
16         }
17         
18         return index+1;
19     }
20 };

Array-Remove Duplicates from Sorted Array

原文：http://www.cnblogs.com/gatsbydhn/p/4970366.html