lintcode: Partition Equal Subset Sum_canpartition lintcode-优快云博客

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。通过对示例的分析和提供一种动态规划解决方案，本文详细解释了实现这一目标的具体步骤。

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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.

Notice

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

Example

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

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

class Solution {
public:
    /**
     * @param nums a non-empty array only positive integers
     * @return return true if can partition or false
     */
    bool canPartition(vector<int>& nums) {
            // Write your code here
            
    	int sum = 0;
    	for (int i = 0; i< nums.size(); i++)
    		sum += nums[i];
    
    	if (sum % 2)
    		return false;
    
    	int target = sum / 2;
    
    	vector<vector<bool>> dp(nums.size() + 1, vector<bool>(target + 1));
    
    	for (int i = 0; i <= nums.size(); i++)
    	{
    		for (int j = 0; j <= target; j++)
    		{
    			dp[i][j] = false;
    		}
    	}
    
    	for (int i = 0; i <= nums.size(); i++)
    		dp[i][0] = true;
    
    
    	for (int i = 1; i <= nums.size(); i++)
    	{
    		for (int k = 1; k <= target; k++)
    		{
    			if (dp[i - 1][k] == true)
    			{
    				dp[i][k] = true;
    			}
    
    			if (k - nums[i - 1] >= 0 && dp[i - 1][k - nums[i - 1]] == true)
    			{
    				dp[i][k] = true;
    			}
    		}
    	}
    
    	return dp[nums.size()][target];
        
    }
};