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219.contains-duplicate-ii.md

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题目地址

https://leetcode.com/problems/contains-duplicate-ii/description/

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 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

思路

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。

关键点解析

代码

  • 语言支持:JS,Python,C++

Javascript Code:

/*
 * @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
};

Python Code:

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

C++ Code:

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