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 的差的绝对值最大为 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
来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/contains-duplicate-ii 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
由于题目没有对空间复杂度有求,用一个hashmap 存储已经访问过的数字即可, 每次访问都会看hashmap中是否有这个元素,有的话拿出索引进行比对,是否满足条件(相隔不大于k),如果满足返回true即可。
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/*
* @lc app=leetcode id=219 lang=javascript
*
* [219] Contains Duplicate II
*
* https://leetcode.com/problems/contains-duplicate-ii/description/
*
* algorithms
* Easy (34.75%)
* Total Accepted: 187.3K
* Total Submissions: 537.5K
* Testcase Example: '[1,2,3,1]\n3'
*
* 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
* absolute difference between i and j is at most k.
*
*
* Example 1:
*
*
* Input: nums = [1,2,3,1], k = 3
* Output: true
*
*
*
* Example 2:
*
*
* Input: nums = [1,0,1,1], k = 1
* Output: true
*
*
*
* Example 3:
*
*
* Input: nums = [1,2,3,1,2,3], k = 2
* Output: false
*
*
*
*
*
*/
/**
* @param {number[]} nums
* @param {number} k
* @return {boolean}
*/
var containsNearbyDuplicate = function(nums, k) {
const visited = {};
for(let i = 0; i < nums.length; i++) {
const num = nums[i];
if (visited[num] !== undefined && i - visited[num] <= k) {
return true;
}
visited[num] = i;
}
return false
};