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【一天一道LeetCode】#26. Remove Duplicates from Sorted Array

时间:2016-04-26 20:46:36      阅读:108      评论:0      收藏:0      [点我收藏+]

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一天一道LeetCode系列

(一)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.

(二)解题


/*

解题:排好序的数据,删除里面重复的数据

需要注意以下两点:

1.erase()调用之后迭代器失效,需要将iter = nums.erase(iter);

2.考虑nums为空或者只有1个的情况,可以直接返回

*/

class Solution {

public:

    int removeDuplicates(vector<int>& nums) {

        if(nums.size()<=1) return nums.size();

        auto iter = nums.begin() +1;

        for(;iter!=nums.end();)

        {

            if(*iter == *(iter-1))

            {

                iter = nums.erase(iter);//关键!erase()返回的是删除的数据的下一个迭代器

            }

            else

                ++iter;//没有删除元素的时候+1

        }

        return nums.size();

    }

};

【一天一道LeetCode】#26. Remove Duplicates from Sorted Array

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原文地址:http://blog.csdn.net/terence1212/article/details/51242161

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