416. Partition Equal Subset Sum

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:

Each of the array element will not exceed 100.
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.

题目含义：多个正数组成的数组，能否分成两个子数组，使得两个子数组的和相等

 1     public boolean canPartition(int[] nums) {
 2         int sum=0;
 3         for (int num:nums) sum+= num;
 4         if(sum % 2 == 1) return false;
 5         sum /=2;
 6         boolean[] dp = new boolean[sum + 1];//当前可以由考虑了的元素构成的值（每个元素最多出现1次）
 7         //0可以由给定的nuns数组的元素组成给出
 8         dp[0] = true;
 9         int curSum = 0;
10         //nums中的元素可以划分为2类：考虑了的和没考虑了的
11         for (int num: nums) {//一个一个元素按顺序加入考虑了的划分中，从而考虑全部可以构成的数的值
12             int tmax = Math.min(curSum + num, sum);//内循环的上限的有效值=min（当前考虑了的划分中的元素可以构成的最大的值，sum）
13             for (int j = tmax; j >= num; j--) {
14                 //构成j值只有2种情况：包含num，不包含num
15                 //包含nums[i]：dp[j] = dp[j - num]
16                 //不包含nums[i]：dp[j] = dp[j]
17                 dp[j] = dp[j] || dp[j - num];
18             }
19             if (dp[sum]) return true;//一旦满足直接返回，减少遍历
20             curSum += nums[i];
21         }
22         return dp[sum]; 
23     }