E

方法1: No brain: HashSet. O(n), 但是实际上比方法2 要慢.

方法2: 排序, 重复数会排在一起. Arrays.sort() time complexity nLog(n)

```

/*

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.

*/

/*

Use a hashSet. Cost: extra space.

*/

class Solution {

public boolean containsDuplicate(int[] nums) {

if (nums == null || nums.length == 0) {

return false;

}

final HashSet<Integer> set = new HashSet<>();

for (int num : nums) {

if (set.contains(num)) {

return true;

}

set.add(num);

}

return false;

}

}

/*

Sorry array, nLog(n)

*/

class Solution {

public boolean containsDuplicate(int[] nums) {

if (nums == null || nums.length == 0) {

return false;

}

Arrays.sort(nums);

for (int i = 0; i < nums.length - 1; i++) {

if (nums[i] == nums[i + 1]) {

return true;

}

}

return false;

}

}

```