416. Partition Equal Subset Sum

来源：互联网发布：淘宝快递拒签运费谁负编辑：程序博客网时间：2024/04/30 14:28

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.

这是原本的思路，与题40类似，但是时间超了。

public class Solution {    public boolean canPartition(int[] nums) {        int k=0;        for(int t:nums){            k+=t;        }        if(k%2!=0) return false;        Arrays.sort(nums);        LinkedList<Integer> list=new LinkedList<>();        helper(nums,k/2,0,list);        return list.size()!=0;    }    public void helper(int[] nums,int k,int index,List<Integer> flag){        if(flag.size()!=0||k<0) return;        if(k==0){            flag.add(1);            return;        }        for(int i=index;i<nums.length;i++){            if(flag.size()!=0)                break;            helper(nums,k-nums[i],i+1,flag);        }    }}

https://discuss.leetcode.com/topic/67539/0-1-knapsack-detailed-explanation

二维数组可以再降维。

public class Solution {    public boolean canPartition(int[] nums) {        // check edge case        if (nums == null || nums.length == 0) {            return true;        }        // preprocess        int volumn = 0;        for (int num : nums) {            volumn += num;        }        if (volumn % 2 != 0) {            return false;        }        volumn /= 2;        // dp def        boolean[] dp = new boolean[volumn + 1];        // dp init        dp[0] = true;        // dp transition        for (int i = 1; i <= nums.length; i++) {            for (int j = volumn; j >= nums[i-1]; j--) {                dp[j] = dp[j] || dp[j - nums[i-1]];            }        }        return dp[volumn];    }}

0 0