leetcode 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.

首先将所有元素相加，若为奇数直接返回false，为偶数，除以2得到需要组合得到的加和。目的就是从原始数组中找到这样一个组合。这就是一个背包问题
暴力解法，对于每个元素有加与不加两种选择，递归的调用，这样照成的复杂度为$O(2^N)$。暴力解法如下：

    boolean isValid(int[] nums, int target, int i){
        if(i==nums.length) return false;
        if(nums[i]==target) return true;
        return isValid(nums, target - nums[i], i+1) || isValid(nums, target, i+1);
    }

采用动态规划的方法
$d[i][j]$表示用数组的前i个元素能否组成j
如果不选第i个元素，那么$d[i][j]=d[i-1][j]$，否则$d[i][j] = d[i-1][j-nums[i]]$，即前i-1个元素能否组成$j-nums[i]$，这样解法为

public boolean canPartition(int[] nums) {
    int sum = 0;

    for (int num : nums) {
        sum += num;
    }

    if ((sum & 1) == 1) {
        return false;
    }
    sum /= 2;

    int n = nums.length;
    boolean[][] dp = new boolean[n+1][sum+1];
    for (int i = 0; i < dp.length; i++) {
        Arrays.fill(dp[i], false);
    }

    dp[0][0] = true;

    for (int i = 1; i < n+1; i++) {
        dp[i][0] = true;
    }
    for (int j = 1; j < sum+1; j++) {
        dp[0][j] = false;
    }

    for (int i = 1; i < n+1; i++) {
        for (int j = 1; j < sum+1; j++) {
            dp[i][j] = dp[i-1][j];
            if (j >= nums[i-1]) {
                dp[i][j] = (dp[i][j] || dp[i-1][j-nums[i-1]]);
            }
        }
    }

    return dp[n][sum];
}

采用空间压缩的方法可以让dp降为一维

    public boolean canPartition(int[] nums) {
        int sum = 0;
        for(int i=0; i < nums.length; i++){
            sum += nums[i];
        }
        if(sum%2==1) return false;
        sum /=2;
        // return isValid(nums, sum_all, 0);
        boolean[] dp = new boolean[sum+1];
        dp[0] = true;
        for(int i=0; i< nums.length; i++){
            for(int j=sum; j>=nums[i]; j--){
                dp[j] = dp[j] || dp[j-nums[i]];
            }
        }
        return dp[sum];
    }