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LeetCode - 217. Contains Duplicate
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題目: 217. Contains Duplicate
給一數字陣列，找出裡面是否有重複的數字。
解題思路
思路一: iterate陣列nums，並把看過的num存在set中；如果num已經在set中，則返回True； 否則iterate結束(代表沒有找到有重複的num)則返回False。
思路二: 排序陣列nums後，從第二個(index = 1)開始iterate整個陣列，並檢查前一個num與當前num是否相同，相同則重複； 否則iterate結束(代表沒有找到有重複的num)則返回False。
Code
Java:
Runtime 5 ms(96.38%), Memory 50.6 MB(96.43%)
class Solution { public boolean containsDuplicate(int[] nums) { HashSet<Integer> viewed = new HashSet<Integer>(); for(int i=0; i<nums.length; i++){ if(!viewed.add(nums[i])){ return true; } } return false; }}Python3:
Runtime 464 ms(85.88%), Memory 25.9 MB(91.76%)
class Solution: def containsDuplicate(self, nums: List[int]) -> bool: viewed = set() for i in range(len(nums)): if nums[i] in viewed: return True else: viewed.add(nums[i]) return FalseC#:
Runtime 199 ms(66.51%), Memory 52.2 MB(36.17%)
public class Solution { public bool ContainsDuplicate(int[] nums) { HashSet<int> viewed = new HashSet<int>(); for(int i=0; i<nums.Length; i++){ if(!viewed.Add(nums[i])){ return true; } } return false; }}Golang:
Runtime 65 ms(97.15%), Memory 8.9 MB(55.31%)
type void struct{}var empty voidfunc containsDuplicate(nums []int) bool { viewed := make(map[int]void); for _, num := range nums { if _, ok := viewed[num]; ok { return true; }else{ viewed[num] = empty; } } return false;}C++:
Runtime 112 ms(68.51%), Memory 51.5 MB(52.15%)
class Solution {public: bool containsDuplicate(vector<int>& nums) { unordered_set<int> viewed; for(int i=0; i<nums.size(); i++){ if (viewed.count(nums[i])){ return true; }else{ viewed.insert(nums[i]); } } return false; }};C:
Runtime 125 ms(73.38%), Memory 12.6 MB(90.14%)
int compare(const int *a, const int *b){ return *a - *b;}bool containsDuplicate(int* nums, int numsSize){ qsort(nums, numsSize, sizeof(int), compare); for(int i=1; i<numsSize; i++){ if (nums[i - 1] == nums[i]){ return true; } } return false;}