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.

将第i的不重复的元素复制到nums[i],然后移除后续元素。时间:32ms

代码如下:

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

 

posted on 2015-06-03 23:01  NealCaffrey989  阅读(91)  评论(0编辑  收藏  举报