217. Contains Duplicate【easy】

217. Contains Duplicate【easy】

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.

解法一：

 1 class Solution {
 2 public:
 3     bool containsDuplicate(vector<int>& nums) {
 4         if (nums.empty()) {
 5             return false;
 6         }
 7         
 8         unordered_map<int, int> my_map;
 9         for (int i = 0; i < nums.size(); ++i) {
10             if (++my_map[nums[i]] > 1) {
11                 return true;
12             }
13         }
14         
15         return false;
16     }
17 };

解法二：

1 class Solution {
2 public:
3     bool containsDuplicate(vector<int>& nums) {
4         return nums.size() > set<int>(nums.begin(), nums.end()).size();        
5     }
6 };

参考@chammika 的代码

解法三：

 1 public boolean containsDuplicate(int[] nums) {
 2     for(int i = 0; i < nums.length; i++) {
 3         for(int j = i + 1; j < nums.length; j++) {
 4             if(nums[i] == nums[j]) {
 5                 return true;
 6             }
 7         }
 8      }
 9     return false;
10 }

Time complexity: O(N^2), memory: O(1)

The naive approach would be to run a iteration for each element and see whether a duplicate value can be found: this results in O(N^2) time complexity.

解法四：

1 public boolean containsDuplicate(int[] nums) {
2     Arrays.sort(nums);
3     for(int ind = 1; ind < nums.length; ind++) {
4         if(nums[ind] == nums[ind - 1]) {
5             return true;
6         }
7     }
8     return false;
9 }

Time complexity: O(N lg N), memory: O(1) - not counting the memory used by sort

Since it is trivial task to find duplicates in sorted array, we can sort it as the first step of the algorithm and then search for consecutive duplicates.

解法五：

 1 public boolean containsDuplicate(int[] nums) {
 2 
 3     final Set<Integer> distinct = new HashSet<Integer>();
 4     for(int num : nums) {
 5         if(distinct.contains(num)) {
 6             return true;
 7         }
 8         distinct.add(num);
 9     }
10     return false;
11 }

Time complexity: O(N), memory: O(N)

Finally we can used a well known data structure hash table that will help us to identify whether an element has been previously encountered in the array.

解法三、四、五均参考@jmnarloch 的代码