containsDuplicate精品文章

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  LeetCode-containsDuplicate-存在重復(fù)

  ...，但是題目提示超過(guò)時(shí)間限制了 class Solution(object): def containsDuplicate(self, nums): :type nums: List[int] :rtype: bool if len(nums)==0: return False ...

  魏明 2019-07-30 16:09 評(píng)論0 收藏0
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  leetcode217.219.220 contains duplicate

  ...起來(lái)，然后查詢當(dāng)前的值是否已經(jīng)便利過(guò) public boolean containsDuplicate(int[] nums) { Map map = new HashMap(); for(int curNum : nums){ if(map.containsKey(curNum)) return true; ...

  tulayang 2019-08-16 10:56 評(píng)論0 收藏0
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  [LeetCode] Contains Duplicate

  ...ment is distinct. Solution Use HashSet class Solution { public boolean containsDuplicate(int[] nums) { if (nums == null || nums.length == 0) return false; HashSet set = new Hash...

  褰辯話 2019-08-19 10:43 評(píng)論0 收藏0
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  【LeetCode】初級(jí)算法-存在重復(fù)

  ...avaScript實(shí)現(xiàn) /** * @param {number[]} nums * @return {boolean} */ var containsDuplicate = function(nums) { nums.sort(); for(let i = 0; i < nums.length-1; i++){ if(nums[i] == nums[i...

  DrizzleX 2019-08-23 11:43 評(píng)論0 收藏0
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  leetcode 217 Contains Duplicate

  ...臨的元素是否相等就可以了 解法一 HashMap public boolean containsDuplicate(int[] nums) { int length = nums.length; if(length

  philadelphia 2019-08-15 13:53 評(píng)論0 收藏0
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  [Leetcode] Contains Duplicate 包含重復(fù)

  ...，則說(shuō)明有重復(fù)。 代碼 public class Solution { public boolean containsDuplicate(int[] nums) { Set set = new HashSet(); for(int i = 0; i < nums.length; i++){ if(set.contains(num...

  rozbo 2019-08-14 12:28 評(píng)論0 收藏0
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  javascript數(shù)組的常用算法解析

  ...否存在重復(fù) /** * @param {number[]} nums * @return {boolean} */ var containsDuplicate = function(nums) { let hashMap = new Map(); for(let i = 0; i < nums.length; i++) { i...

  xumenger 2019-08-23 16:04 評(píng)論0 收藏0
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  TypeScript實(shí)現(xiàn)數(shù)組相關(guān)簡(jiǎn)單算法

  ...相同，則返回 false。 * * 這個(gè)一定不是最優(yōu)解 */ const containsDuplicate = function (nums: number[]) :boolean{ // 設(shè)置flag let judge = false // 容錯(cuò)判斷 if (nums.length { //每次循環(huán)小游標(biāo) let count =...

  cloud 2019-08-23 11:55 評(píng)論0 收藏0