Partition Equal Subset Sum

描述

Given a non-empty array nums 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.

Example 1:

Input: nums = [1,5,11,5] > Output: true Explanation: The array can be partitioned as [1, 5, 5] and [11].

Example 2:

Input: nums = [1,2,3,5] > Output: false Explanation: The array cannot be partitioned into equal sum subsets.

Constraints:

• 1 <= nums.length <= 200
• 1 <= nums[i] <= 100

分析

本题是一个0-1背包，且背包恰好装满的问题。令每个物品$i$的重量$w_i$为 nums[i]，价值$v_i$为 0，背包能容纳的最大重量$W=\frac{1}{2}\sum_i w_i$，该问题就变成，选择若干物品，能否恰好填满背包？

令 f(i, j)表示前 $i$ 个物品能否填满容量为 $j$ 的背包，则状态转移方程为:

$f(i,j) = f(i-1,j) \lor f(i-1, j-w_i)$

代码

Python

# TODO

Java

// Partition Equal Subset Sum
// 0-1 knapsack problem
// Time Complexity: O(n*W), Space Complexity: O(W)
class Solution {
    public boolean canPartition(int[] nums) {
        int sum = 0;
        for(int i : nums) sum += i;
        if(sum % 2 != 0) return false;

        int[] w = nums; // weight array
        int W = sum / 2; // maximum weight capacity of knapsack

        boolean[] dp = new boolean[W + 1];
        dp[0] = true; // initialize

        for(int i = 0; i < nums.length; i++) {
            for(int j = W; j >= w[i]; --j) {
                dp[j] = dp[j] || dp[j-w[i]];
            }
        }
        return dp[W];
    }
}

C++

// Partition Equal Subset Sum
// 0-1 knapsack problem
// Time Complexity: O(n*W), Space Complexity: O(W)
class Solution {
public:
    bool canPartition(vector<int>& nums) {
        int sum = 0;
        for(int i : nums) sum += i;
        if(sum % 2 != 0) return false;

        const vector<int>& w = nums; // weight array
        int W = sum / 2; // maximum weight capacity of knapsack

        vector<bool> dp(W + 1);
        dp[0] = true; // base case

        for(int i = 0; i < nums.size(); i++) {
            for(int j = W; j >= w[i]; --j) {
                dp[j] = dp[j] || dp[j-w[i]];
            }
        }
        return dp[W];
    }
};

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