lintcode(588)Partition Equal Subset Sum

最新推荐文章于 2025-08-20 17:55:57 发布

sunday0904

最新推荐文章于 2025-08-20 17:55:57 发布

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文章标签： java lintcode

本文链接：https://blog.youkuaiyun.com/sunday0904/article/details/72571133

本文探讨了一个经典的计算机科学问题——如何判断一个只包含正整数的非空数组能否被划分为两个子集，使得这两个子集的元素之和相等。文章提供了一种解决方案，通过计算数组元素总和的一半作为目标值，并将问题转化为背包问题来解决。

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Description:

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.

注意事项

Each of the array element will not exceed 100.
The array size will not exceed 200.

Explanation:

Given nums = [1, 5, 11, 5], return true
two subsets: [1, 5, 5], [11]

Given nums = [1, 2, 3, 9], return false

Solution:

First, calculate the sum of nuts. If the result is odd, return false. If not, set the target as half of the sum.Then we can transform the problem to backpack.

Regard mums[i] as element, target as the capacity of the backpack.

public class Solution {
    /**
     * @param nums a non-empty array only positive integers
     * @return return true if can partition or false
     */
    public boolean canPartition(int[] nums) {
        // Write your code here
        int sum = 0;
        int len = nums.length;
        for(int i = 0 ; i < len ; i++){
            sum += nums[i];
        }
        
        if(sum%2 == 1) return false;
        int target = sum/2;
        
        int[][] dp = new int[len][target + 1];
        
        for(int i = nums[0];i<=target;i++){
            dp[0][i] = nums[0];
        }
        
        for(int i = 1;i<len;i++){
            for(int j = nums[i];j<=target;j++){
                dp[i][j] = Math.max(dp[i-1][j] , dp[i - 1][j - nums[i]] + nums[i]);
            }
        }
        
        if(dp[len - 1][target] == target) return true;
        return false;
    }
}