698. Partition to K Equal Sum Subsets

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问题描述:

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.

 

解题思路:

首先我们要确定当前数组的和能分整除k,若不能整除k,则直接返回false。若能够整除,则可以进一步检查能否分成这些部分。

可以用dfs来解答。

需要用一个visited数组来记录是否已经用过这个数字。

用k来表示还剩多少部分, start 表示从哪里开始找,target表示目标和,cur_sum表示当前和。

 

代码:

DFS解法:

class Solution {
public:
    bool canPartitionKSubsets(vector<int>& nums, int k) {
        if(nums.empty()) return false;
        int sum = 0;
        for(int i : nums){
            sum += i;
        }
        if(sum % k != 0) return false;
       
        vector<bool> visited(nums.size(), false);
        return canPartition(nums, visited, 0, k, 0, sum/k);
    }
    
    bool canPartition(vector<int> &nums, vector<bool> &visited, int start, int k, int cur_sum, int target){
        if(k == 1) return true;
        if(cur_sum == target) return canPartition(nums, visited, 0, k-1, 0, target);
        for(int i = start; i < nums.size(); i++){
            if(!visited[i]){
                visited[i] = true;
                if(canPartition(nums,visited, i+1, k, cur_sum+nums[i], target)) return true;
                visited[i] = false;
            }
        }
        return false;
    }
};

 

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