Partition Equal Subset Sum (M)

題目

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

題意

在一個給定陣列中找到一個子陣列，使其和為指定值。

思路

使用記憶化搜尋容易實現。

程式碼實現

Java

class Solution {
    public boolean canPartition(int[] nums) {
        int total = 0;
        int left = 0, right = nums.length - 1;
        
        for (int num : nums) {
            total += num;
        }

        if (total % 2 == 1) {
            return false;
        }
        
        return dfs(total / 2, nums, 0, new Boolean[total / 2 + 1][nums.length]);
    }
    
    private boolean dfs(int sum, int[] nums, int start, Boolean[][] record) {
        if (start == nums.length) {
            return sum == 0;
        }
        if (sum < 0) {
            return false;
        }
        if (record[sum][start] != null) {
            return record[sum][start];
        }
        
        boolean found = dfs(sum, nums, start + 1, record) || dfs(sum - nums[start], nums, start + 1, record);
        record[sum][start] = found;
        return found;
    }
}
```# Partition Equal Subset Sum (M)

## 題目

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`

---

## 題意

在一個給定陣列中找到一個子陣列，使其和為指定值。

## 思路

使用記憶化搜尋容易實現。

---

## 程式碼實現

### Java

```java
class Solution {
    public boolean canPartition(int[] nums) {
        int total = 0;
        int left = 0, right = nums.length - 1;
        
        for (int num : nums) {
            total += num;
        }

        if (total % 2 == 1) {
            return false;
        }
        
        return dfs(total / 2, nums, 0, new Boolean[total / 2 + 1][nums.length]);
    }
    
    private boolean dfs(int sum, int[] nums, int start, Boolean[][] record) {
        if (start == nums.length) {
            return sum == 0;
        }
        if (sum < 0) {
            return false;
        }
        if (record[sum][start] != null) {
            return record[sum][start];
        }
        
        boolean found = dfs(sum, nums, start + 1, record) || dfs(sum - nums[start], nums, start + 1, record);
        record[sum][start] = found;
        return found;
    }
}