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:

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.

分析：https://leetcode.com/problems/partition-equal-subset-sum/discuss/90592/01-knapsack-detailed-explanation

This problem is essentially let us to find whether there are several numbers in a set which are able to sum to a specific value (in this problem, the value is sum/2).

Actually, this is a 0/1 knapsack problem, for each number, we can pick it or not. Let us assume dp[i][j] means whether the specific sum j can be gotten from the first i numbers. If we can pick such a series of numbers from 0-i whose sum is j, dp[i][j] is true, otherwise it is false.

Base case: dp[0][0] is true; (zero number consists of sum 0 is true)

Transition function: For ith number, if we don‘t pick it, dp[i][j] = dp[i-1][j], which means if the first i-1 elements has made it to j, dp[i][j] would also make it to j (we can just ignore nums[i]). If we pick nums[i]. dp[i][j] = dp[i-1][j-nums[i]], which represents that j is composed of the current value nums[i] and the remaining composed of other previous numbers. Thus, the transition function is dp[i][j] = dp[i-1][j] || dp[i-1][j-nums[i]]

 1 class Solution {
 2     public boolean canPartition(int[] nums) {
 3         int sum = 0;
 4         for (int num : nums) {
 5             sum += num;
 6         }
 7
 8         if ((sum & 1) == 1) {
 9             return false;
10         }
11         sum /= 2;
12
13         int n = nums.length;
14         boolean[][] dp = new boolean[n + 1][sum + 1];
15
16         dp[0][0] = true;
17
18         for (int i = 1; i < n + 1; i++) {
19             dp[i][0] = true;
20         }
21         for (int j = 1; j < sum + 1; j++) {
22             dp[0][j] = false;
23         }
24
25         for (int i = 1; i <= n; i++) {
26             for (int j = 1; j <= sum; j++) {
27                 dp[i][j] = dp[i - 1][j];
28                 if (j >= nums[i - 1]) {
29                     dp[i][j] = dp[i][j] || dp[i - 1][j - nums[i - 1]];
30                 }
31             }
32         }
33
34         return dp[n][sum];
35     }
36 }

原文地址：https://www.cnblogs.com/beiyeqingteng/p/10817311.html