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%)
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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;
}
}
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Python3:
Runtime 464 ms(85.88%), Memory 25.9 MB(91.76%)
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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 False
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C#:
Runtime 199 ms(66.51%), Memory 52.2 MB(36.17%)
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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;
}
}
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Golang:
Runtime 65 ms(97.15%), Memory 8.9 MB(55.31%)
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type void struct{}
var empty void
func 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;
}
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C++:
Runtime 112 ms(68.51%), Memory 51.5 MB(52.15%)
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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;
}
};
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C:
Runtime 125 ms(73.38%), Memory 12.6 MB(90.14%)
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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;
}
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