LeetCode 26

一、问题描述

Description:

Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.

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.

Note:

Do not allocate extra space for another array, you must do this in place with constant memory.

给一个有序数组，原地移除重复的元素，并返回新数组的长度。

注意：只能原地工作，不能使用额外的数组。

二、解题报告

由于输入的是有序数组，在遍历过程中，我们只需要判断是否与前面的重复即可。

代码：

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if(nums.empty()) 
            return 0;
        int count = 1;
        int pre = nums[0];
        vector<int>::iterator beg = nums.begin();
        for(++beg; beg!=nums.end(); ) {   // 从第二个开始
            if(*beg == pre)               // 重复，移除
                beg = nums.erase(beg);
            else                          // 不重复，更新pre
            {
                pre = *beg;
                ++beg;
            }
        }
        return nums.size();
    }
};

LeetCode答案源代码：https://github.com/SongLee24/LeetCode