https://leetcode.com/problems/contains-duplicate-ii/description/
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
- hashmap
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。
无
- 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;
}
};