Leetcode 217 - Contains Duplicate

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.

需要能够快速查询，然后有一个数和对应出现次数的对应，所以想到用Hashtable
Solution1:

import java.util.*;
public class Solution {
    public boolean containsDuplicate(int[] nums) {
        Hashtable nummap = new Hashtable();
        for (int n : nums) {
            if (nummap.get(n) != null) return true;
            else nummap.put(n,1);
        }
        return false;
    }
}

其实只关心每个num出现没出现，0或者1，所以用不到map，只需要知道有没有就可以，所以hashset就可以。
Solution2:

public class Solution {
    public boolean containsDuplicate(int[] nums) {
        HashSet numset = new HashSet();
        for (int n : nums) {
            if (!numset.add(n)) return true;
        }
        return false;
    }
}

Leetcode 238 - Product of Array Except Self

Given an array of n integers where n > 1, nums, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].

Solve it without division and in O(n).

For example, given [1,2,3,4], return [24,12,8,6].

Follow up:
Could you solve it with constant space complexity? (Note: The output array does not count as extra space for the purpose of space complexity analysis.)
O(n), 所以只能扫一遍就得到该有的信息，自然想到的是左边扫一遍右边扫一遍。
省空间的话简单，右边扫的时候直接把答案存起来。

public class Solution {
    public int[] productExceptSelf(int[] nums) {
        //keep track of product of left elements in the array
        // then scan from right to left to multiply the right elements
        int n = nums.length;
        int[] results = new int[n];
        results[0] = 1;
        for (int i = 1; i < n; i++) {
            results[i] = results[i-1] * nums[i-1];
        }
        int right = 1;
        for (int i = n-1; i >= 0; i--) {
            results[i] *= right;
            right *= nums[i];
        }
        return results;
    }
}