[LeetCode] 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.

对一个排序好的数组去重，并返回去重后的数组的长度。
利用两个指针来遍历数组，其一为cnt表示非重复元素所在数组中的最后位置。另一个i为顺序遍历数组的位置。每次用cnt与i所指的元素比较，如果不同，则更新cnt所指元素为新非重元素，即i所指的元素。

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if (nums.empty())
            return 0;
        int cnt = 0;
        for (int i = 1; i != nums.size(); i++) {
            if (nums[cnt] != nums[i]) {
                cnt++;
                nums[cnt] = nums[i];
            }
        }
        return cnt + 1;
    }
};
// 29 ms

[LeetCode] Remove Duplicates from Sorted Array

原文：http://www.cnblogs.com/immjc/p/7622901.html