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219. 存在重复元素 II

English Version

题目描述

给定一个整数数组和一个整数 k，判断数组中是否存在两个不同的索引 i 和 j，使得 nums [i] = nums [j]，并且 i 和 j 的差的 绝对值 至多为 k。

 

示例 1:

输入: nums = [1,2,3,1], k = 3
输出: true

示例 2:

输入: nums = [1,0,1,1], k = 1
输出: true

示例 3:

输入: nums = [1,2,3,1,2,3], k = 2
输出: false

解法

Python3

class Solution:
    def containsNearbyDuplicate(self, nums: List[int], k: int) -> bool:
        mp = {}
        for i, v in enumerate(nums):
            if v in mp and i - mp[v] <= k:
                return True
            mp[v] = i
        return False

Java

class Solution {
    public boolean containsNearbyDuplicate(int[] nums, int k) {
        Map<Integer, Integer> mp = new HashMap<>();
        for (int i = 0; i < nums.length; ++i) {
            if (mp.containsKey(nums[i]) && i - mp.get(nums[i]) <= k) {
                return true;
            }
            mp.put(nums[i], i);
        }
        return false;
    }
}

C++

class Solution {
public:
    bool containsNearbyDuplicate(vector<int>& nums, int k) {
        unordered_map<int, int> mp;
        for (int i = 0; i < nums.size(); ++i)
        {
            if (mp.count(nums[i]) && i - mp[nums[i]] <= k) return true;
            mp[nums[i]] = i;
        }
        return false;
    }
};

Go

func containsNearbyDuplicate(nums []int, k int) bool {
	mp := make(map[int]int)
	for i, v := range nums {
		if j, ok := mp[v]; ok {
			if i-j <= k {
				return true
			}
		}
		mp[v] = i
	}
	return false
}

C#

public class Solution {
    public bool ContainsNearbyDuplicate(int[] nums, int k) {
        var mp = new Dictionary<int, int>();
        for (int i = 0; i < nums.Length; ++i)
        {
            if (mp.ContainsKey(nums[i]) && i - mp[nums[i]] <= k)
            {
                return true;
            }
            mp[nums[i]] = i;
        }
        return false;
    }
}

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