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: trueExplanation: The array can be partitioned as [1, 5, 5] and [11].
Example 2:
Input: [1, 2, 3, 5]Output: falseExplanation: 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true10 for (int i = 1; i <= n; i++) dp[i][0] = true;//用i个数的子集可以构成011 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 }