【LeetCode】Contains Duplicate II 解题小结 题目:Given an array of integers and an integerk, find out whether there are two distinct indicesiandjin the array such that nums[i] = nums[j] and the difference betweeniandjis at mostk. 题目意思就是判断在一个数组中是否存在两个相同的元素,他们之...
Output: false Constraints: 1 <= nums.length <= 105 -109 <= nums[i] <= 109 0 <= k <= 105 My Solution: fromcollectionsimportCounterclassSolution(object):defcontainsNearbyDuplicate(self, nums, k):""":type nums: List[int] :type k: int :rtype: bool"""ifk ==0:returnFalse counter=...
因为这里要判断 j−i<=k, 我们只需要使得 hash map的value保存最近的index就行。 见code: class Solution(object): def containsNearbyDuplicate(self, nums, k): """ :type nums: List[int] :type k: int :rtype: bool """ mydict = {} for i in xrange(len(nums)): if nums[i] in mydic...
[LeetCode] Contains Duplicate II 包含重复值之二
Contains Duplicate - Leetcode 217 - Python, 视频播放量 5、弹幕量 0、点赞数 0、投硬币枚数 0、收藏人数 0、转发人数 0, 视频作者 呼吸的chou, 作者简介 :),相关视频:【Python爬虫】手把手教你20行代码永久白嫖VIP付费电影,可分享源码,轻松实现看电影自由!python爬
public class Solution {public bool ContainsNearbyAlmostDuplicate(int[] nums, int k, int t) {var len = nums.Length;var arr = nums.Select(((num, index) => new {num, index})).OrderBy(u => u.num).ToArray();for (int i = 0; i < len; i++)for (int j = i + 1; j < len...
Remove(nums[i - indexDiff]) } } // 在循环中没有返回,则所有数都不满足题意 return false } 题目链接: Contains Duplicate III: leetcode.com/problems/c 存在重复元素 III: leetcode.cn/problems/co LeetCode 日更第 282 天,感谢阅读至此的你 欢迎点赞、收藏、在看鼓励支持小满...
2949.Count-Beautiful-Substrings-II (H-) Heap 220.Contains-Duplicate-III (M) 295.Find-Median-from-Data-Stream (M) 363.Max-Sum-of-Rectangle-No-Larger-Than-K (H) 352.Data-Stream-as-Disjoint-Intervals (H) 480.Sliding-Window-Median (H) 699.Falling-Squares (H) 729.My-Calendar-I (M) ...
今天介绍的是LeetCode算法题中Easy级别的第52题(顺位题号是217)。给定一个整数数组,查找数组是否包含任何重复项。如果数组中至少出现两次值,则函数应返回true,如果每个元素都不相同,则返回false。例如: 输入:[1,2,3,1] 输出:true 输入:[1,2,3,4] 输出:false 输入:[1,1,1,3,3,4,3,2,4,2] 输出:...
// https://discuss.leetcode.com/topic/15305/simple-java-solution/6publicbooleancontainsNearbyDuplicate(int[]nums,intk){Set<Integer>set=newHashSet<Integer>();for(inti=0;i<nums.length;i++){if(i>k)set.remove(nums[i-k-1]);if(!set.add(nums[i]))returntrue;}returnfalse;}...