解法二:排序 class Solution: ## 先排序,后逐一对比。NlogN--1 def containsDuplicate(self, nums: List[int]) -> bool: ## 补缺 if len(nums)<=1: return False nums.sort() # Timsort in Python, nlogn for i in range(len(nums)): if nums[i]==nums[i-1]: return True return False ...
代码如下: classSolution {public:boolcontainsNearbyDuplicate(vector<int>& nums,intk) { /* 声明一个unordered_map m; */ unordered_map<int,int>m ; /* 遍历数组 */for(inti =0; i < nums.size(); ++i) { /* 查找m中是否存在这个元素,若不在,则插入这个元素; 若这个元素在m中,则判断当前 的...
LeetCode:Contains Duplicate Problem: Given an array of integers, find if the array contains any duplicates. Your function should return true if any value appears at least twice in the array, and it should return false if every element is distinct. Solution:用hashmap来计数判断重复次数 1 2 3 ...
解法代码: ## LeetCode 217EfromtypingimportListclassSolution:defcontainsDuplicate(self,nums:List[int])->bool:len1=len(nums)iflen1<=1:returnFalse## 如果长度小于等于1,就不可能出现重复## 排序nums.sort()foriinrange(len1-1):## 比较 len-1 趟ifnums[i]==nums[i+1]:## 排序之后,相同的元素...
LeetCode Contains Duplicate (判断重复元素) 题意: 如果所给序列的元素不是唯一的,则返回true,否则false。 思路: 哈希map解决。 1 class Solution { 2 public: 3 bool containsDuplicate(vector<int>& nums) { 4 unordered_map<int,int> mapp; 5 for(int i=0; i<nums.size(); i++)...
class Solution(object): def containsDuplicate(self, nums): return len(set(nums)) != len(nums) if len(nums) != 0 else False 1. 2. 3.
class Solution(object):def containsNearbyAlmostDuplicate(self, A, k, t):""":type A: List[int]:type k: int:type t: int:rtype: bool"""n = len(A)A = zip(A, range(n))A.sort()for i in xrange(n):j = i + 1while j < n and A[j][0] - A[i][0] <= t:if abs(A[j...
classSolution:defcontainsDuplicate(self,nums:List[int])->bool:ifnot nums orlen(nums)<1:returnFalse #set''' len1=len(nums)set1=set(nums)len2=len(set1)iflen2<len1:returnTruereturnFalse''' # 哈希表''' res={}fornuminnums:ifnuminres:returnTrue ...
classSolution {public:boolcontainsNearbyDuplicate(vector<int>& nums,intk) { unordered_map<int,int>m;for(inti =0; i < nums.size(); ++i) {if(m.find(nums[i]) != m.end() && i - m[nums[i]] <= k)returntrue;elsem[nums[i]] =i; ...
classSolution { public: boolcontainsDuplicate(vector<int>& nums) { unordered_map<int,int> myMap; for(inti = 0;i < nums.size();i++) { if(myMap.find(nums[i]) == myMap.end() ) { myMap.insert(pair<int,int>(nums[i],1)); ...