LeetCode 416. 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：

其实是找有没有sub array的数字和是sum/2. Thus for the other half, sum must be sum/2.

Let dp[i] denotes if there is subarray sum equal to j.

dp[0] = true. 

Then for each num in nums, dp[num] would be true.

Thus for i, check if dp[i-num], if dp[i-num] is true. Then dp[i] msut be true. 

存储到第 i 个数能不能合成 j. 

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:

class Solution {
    public boolean canPartition(int[] nums) {
        if(nums == null || nums.length == 0){
            return true;
        }
        
        int sum = 0;
        for(int num : nums){
            sum += num;
        }
        
        if(sum % 2 != 0){
            return false;
        }
        
        
        int target = sum/2;
        boolean [] dp = new boolean[target+1];
        dp[0] = true;
        for(int num : nums){
            for(int i = target; i>=num; i--){
                dp[i] = dp[i] || dp[i-num];
            }
        }
        
        return dp[target];
    }
}

类似Target Sum 的Method 2.

跟上Partition to K Equal Sum Subsets.