[LeetCode] (medium) 416. Partition Equal Subset Sum

本文探讨了非空正整数数组能否被划分为两个子集,使得两子集元素之和相等的问题。通过示例说明了算法实现,包括01背包算法和深度优先搜索(DFS)剪枝策略,展示了如何在LeetCode上解决Partition Equal Subset Sum题目。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

https://leetcode.com/problems/partition-equal-subset-sum/

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:

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

 

Example 1:

Input: [1, 5, 11, 5]
Output: true

Explanation: The array can be partitioned as [1, 5, 5] and [11].

 

Example 2:

Input: [1, 2, 3, 5]
Output: false

Explanation: The array cannot be partitioned into equal sum subsets.

01背包:

class Solution {
public:
    bool canPartition(vector<int>& nums) {
        int total = accumulate(nums.begin(), nums.end(), 0);
        if(total&1) return false;
        
        int target = (total >> 1);
        
        vector<bool> bag(target+1, false);
        
        bag[0] = true;
        
        int last = 0;
        
        for(int cur : nums){
            for(int i = min(target, last+cur); i >= cur; --i){
                if(bag[i-cur]){
                    bag[i] = true;
                    last = max(last, i);
                }
            }
            if(bag[target]) return true;
        }
        
        return bag[target];
    }
};

 

DFS剪枝:

越早剪枝越好

class Solution {
public:
    
    int total;
    int found = false;
    
    bool canPartition(vector<int>& nums) {
        total = accumulate(nums.begin(), nums.end(), 0);
        if(total%2) return false;
        
        total >>= 1;
        
        sort(nums.begin(), nums.end(), std::greater<int>());
        
        return dfs(nums, 1, nums[0]);
    }
    
    bool dfs(vector<int>& nums, int cur, int num){
        if(num == total){
            return true;
        }
        if(cur >= nums.size() || num > total) return false;
        
        return dfs(nums, cur+1, num+nums[cur]) || dfs(nums, cur+1, num);
    }
};

 

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值