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: 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.

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(2N)。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];    }