Leet Code OJ 219. Contains Duplicate II [Difficulty: Easy]

题目:
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 j is at most k.

翻译:
给定一个整数数组和一个整数k。找出是否存在下标i,j使得nums[i] = nums[j]。同一时候i,j的差值小于等于k。

分析:
遍历数组。使用map存储每一个值近期一次出现的下标。

代码:

public class Solution {
    public boolean containsNearbyDuplicate(int[] nums, int k) {
        Map<Integer,Integer> map=new HashMap<>();
        for(int i=0;i<nums.length;i++){
            Integer value=map.get(nums[i]);
            if(value!=null&&i-value<=k){
                return true;
            }else{
                map.put(nums[i],i);
            }
        }
        return false;
    }
}
原文地址:https://www.cnblogs.com/brucemengbm/p/7300439.html