LeetCode Partition Equal Subset Sum

原题链接在这里：https://leetcode.com/problems/partition-equal-subset-sum/description/

题目：

Given a non-empty array containing only positive integers, find if the array can be partitioned into two subsets such that the sum of elements in both subsets is equal.

Note:

1. Each of the array element will not exceed 100.
2. The array size will not exceed 200.

Example 1:

Input: [1, 5, 11, 5]

Output: true

Explanation: The array can be partitioned as [1, 5, 5] and [11].

Example 2:

Input: [1, 2, 3, 5]

Output: false

Explanation: The array cannot be partitioned into equal sum subsets.

题解：

DP问题. 其实是找有没有sub array的数字和是sum(nums)/2. 存储到第 i 个数能不能合成 j.

地推公式dp[i][j] = dp[i-1][j] || dp[i-1][j-nums[i]]. 两种情况，一种是不用当前的数字nums[i]. 就看之前的数字能不能合成j, dp[i-1][j]. 另一种是用当前的数字nums[i], 看之前的数字能不能合成j-nums[i], dp[i-1][j-nums[i]].

初始化dp[i][0]不论用几个数，总能合成0.

优化空间可用一维数组. 但注意循环从后往前，因为新的iteration 会用到旧的iteration 当前位置前面的值.

Time Complexity: O(sum*nums.length). sum = sum(nums)/2.

Space: O(sum).

AC Java:

 1 class Solution {
 2     public boolean canPartition(int[] nums) {
 3         if(nums == null || nums.length == 0){
 4             return true;
 5         }
 6
 7         int sum = 0;
 8         for(int num : nums){
 9             sum += num;
10         }
11
12         return sum%2==1 ? false : subSum(nums, sum/2);
13     }
14
15     private boolean subSum(int [] nums, int sum){
16         boolean [] dp = new boolean[sum+1];
17         dp[0] = true;
18         for(int num : nums){
19             for(int i = sum; i>=num; i--){
20                 dp[i] = dp[i] || dp[i-num];
21             }
22         }
23         return dp[sum];
24     }
25 }

类似Target Sum 的Method 2.