Leetcode 416. Partition Equal Subset Sum

Posted 2020-08-29 Deribs4

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:

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.

 

动态规划，0/1背包问题。对于输入的数组nums，首先nums中各个元素之和要是偶数，然后就是找到可以将nums中的元素划分为2个不相交子集，并且这2个子集互补。

解法一：nums中的元素可以划分为2类：考虑了的和没考虑了的。由大值遍历到小值，具体可以根据代码画图理解。

 1 public class Solution {
 2     public boolean canPartition(int[] nums) {
 3         if (nums == null || nums.length == 0) {
 4             return false;
 5         }
 6         int sum = 0;
 7         for (int num : nums) sum += num;
 8         if (sum % 2 == 1) return false;
 9         sum /= 2;
10         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）
11         //0可以由给定的nuns数组的元素组成给出
12         dp[0] = true;
13         int curSum = 0;
14         //nums中的元素可以划分为2类：考虑了的和没考虑了的
15         for (int i = 0; i < nums.length; i++) {//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
16             int tmax = Math.min(curSum + nums[i], 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）
17             for (int j = tmax; j >= nums[i]; j--) {
18                 //构成j值只有2种情况：包含nums[i]，不包含nums[i]
19                 //包含nums[i]：dp[j] = dp[j - nums[i]]
20                 //不包含nums[i]：dp[j] = dp[j] 
21                 dp[j] = dp[j] || dp[j - nums[i]];
22             }
23             if (dp[sum]) return true;//一旦满足直接返回，减少遍历
24             curSum += nums[i];
25         }
26         return dp[sum];
27     }
28 }

 

解法二：

 1 public class Solution {
 2     public boolean canPartition(int[] nums) {
 3         int sum = 0;
 4         for (int num : nums) sum += num;
 5         if ((sum & 1) == 1) return false;//奇数还是偶数的判断
 6         sum /= 2;
 7         int n = nums.length;
 8         boolean[][] dp = new boolean[n + 1][sum + 1];//dp[i][j]=true表示的能够用前i个数的子集构成j值
 9         dp[0][0] = true;//用0个数构成0是true
10         for (int i = 1; i <= n; i++) dp[i][0] = true;//用i个数的子集可以构成0
11         for (int i = 1; i <= sum; i++) dp[0][i] = false;//用0个数不能构成大于0的任何数
12         for (int i = 1; i <= n ; i++) {
13             for (int j = 1; j <= sum; j++) {
14                 //dp[i][j] = dp[i - 1][j] || dp[i - 1][j - nums[i - 1]]
15                 //也是2种情况：
16                 //包含nums[i - 1]：dp[i][j] = dp[i - 1][j]
17                 //不包含nums[i - 1]：dp[i][j] = dp[i - 1][j - nums[i - 1]]
18                 dp[i][j] = dp[i - 1][j];
19                 if (j >= nums[i - 1]) dp[i][j] = dp[i][j] || dp[i - 1][j - nums[i - 1]];
20             }
21         }
22         return dp[n][sum];
23     }
24 }