leetcode 698. Partition to K Equal Sum Subsets

本文链接：https://blog.youkuaiyun.com/liqinzhe11/article/details/78707113

本文探讨了如何将一组整数等分为k个子集，使每个子集的和相等的问题。通过递归搜索和回溯算法实现解决方案，并提供了一个具体的例子说明其可行性。

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698. Partition to K Equal Sum Subsets

Given an array of integers nums and a positive integer k, find whether it's possible to divide this array into k non-empty subsets whose sums are all equal.

Example 1:

Input: nums = [4, 3, 2, 3, 5, 2, 1], k = 4
Output: True
Explanation: It's possible to divide it into 4 subsets (5), (1, 4), (2,3), (2,3) with equal sums.

Note:

1 <= k <= len(nums) <= 16.

0 < nums[i] < 10000.

class Solution {
public:
    bool canPartitionKSubsets(vector<int>& nums, int k) 
    {
        //先要知道sum是多少
        sum = 0;
        for (auto it : nums)
            sum += it;
        if (sum % k != 0) return false;
        sum = sum / k;
        sort(nums.begin(), nums.end(), greater<int>());
        for (int i = 0; i < k; i++)
            group.push_back(0);
        return helper(nums, 0, group, sum);
    }
    
    bool helper(vector<int>& nums, 
                int i,   //当前需要安排第i个
                vector<int> group,
                int all) //每一组的和
    {
        if (i == nums.size()) return true;
        for (int j = 0; j < group.size(); j++)
        {
            if (group[j] + nums[i] <= all) //可以往里面放
            {
                group[j] += nums[i]; 
                if (helper(nums, i + 1, group, all)) return true;
                group[j] -= nums[i];
            }
        }
        return false;
    }
    
private:
    int sum;
    vector<int> group;
};