<LeetCode OJ> 219. Contains Duplicate II

219. Contains Duplicate II

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Submissions: 142658 Difficulty: Easy

有一个数组和一个整数，如果数组中两个相同的元素（分别在数组中i和j的位置），如果判断是否|i-j|<=k，若是则返回真
Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i]
= nums[j] and the difference between i and jis at most k.

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(E) Contains Duplicate (M)
Contains Duplicate III

//思路首先：哈希map（不要用红黑树map）
//遍历数组，首先看当前元素是否在map中，如不在则压入，若在看是否其对应下标和当前下标相距为k
//如果不则将原元素修改为现在的下标
class Solution {
public:
    bool containsNearbyDuplicate(vector<int>& nums, int k) {
        if(nums.empty())
            return false;
        unordered_map<int,int> umapping;
        umapping[nums[0]]=0;
        for(int i=1;i<nums.size();i++)
        {
            //if(umapping.count(nums[i]) == 0)//没有出现(统计函数)
            if(umapping.find(nums[i]) == umapping.end())//直接查找
                umapping[nums[i]]=i;
            else if((i-umapping[nums[i]])<=k)//相距小于k
                return true;
            else
                umapping[nums[i]]=i;
        }
        return false;
    }
};

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原文地址：http://blog.csdn.net/ebowtang/article/details/50490187

原作者博客：http://blog.csdn.net/ebowtang