LeetCode -- Contains Duplicate II
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题目描述:
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.
在一个数组nums中试着找到两个数nums[i]和nums[j],其中,i与j的距离要小于等于k。如果找到,返回true,否则返回false。
思路:
一次遍历num[i...n),哈希存每个数的位置,如果nums[i]已经出现,就判断上次出现的位置与当前位置的距离是否小于等于k。如果是,返回true;否则,更新Hash[nums[i]]的位置=i。
实现代码:
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.
在一个数组nums中试着找到两个数nums[i]和nums[j],其中,i与j的距离要小于等于k。如果找到,返回true,否则返回false。
思路:
一次遍历num[i...n),哈希存每个数的位置,如果nums[i]已经出现,就判断上次出现的位置与当前位置的距离是否小于等于k。如果是,返回true;否则,更新Hash[nums[i]]的位置=i。
实现代码:
public class Solution { public bool ContainsNearbyDuplicate(int[] nums, int k) { var hash = new Dictionary<int, int>(); for(var i = 0;i < nums.Length; i++){ if(!hash.ContainsKey(nums[i])){ hash.Add(nums[i],i); } else{ if(Math.Abs(hash[nums[i]] - i) <= k){ return true; } else{ hash[nums[i]] = i; } } } return false; }}
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