Leetcode: 698. Partition to K Equal Sum Subsets
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Description
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
用递归实现
1. 求出数组的所有数字之和sum,先判断sum是否能整除k,不能整除的话直接返回false。
2. visited数组来记录哪些数组已经被选中了,然后调用递归函数,目的是组k个子集合,每个子集合之和为target = sum/k。变量start,表示从数组的某个位置开始查找,curSum为当前子集合之和。
3. 在递归函数中,如果k=1,说明此时只需要组一个子集合,那么当前的就是了,直接返回true。
如果curSum等于target了,那么我们再次调用递归,此时传入k-1,start和curSum都重置为0,因当前又找到了一个和为target的子集合,要开始继续找下一个。否则的话就从start开始遍历数组,如果当前数字已经访问过了则直接跳过,否则标记为已访问。
4. 然后调用递归函数,k保持不变,因为还在累加当前的子集合,start传入i+1,curSum传入curSum+nums[i],因为要累加当前的数字,如果递归函数返回true了,则直接返回true。否则就将当前数字重置为未访问的状态继续遍历。
class Solution {public: bool canPartitionKSubsets(vector<int>& nums, int k) { int size = nums.size(); if (size == 0) return false; if (k == 1) return true; if (k < 1 || k > 16 || k > size) return false; int sum = 0; for (int i:nums) { sum += i; } if (sum%k != 0) return false; vector<bool> visited(size,false); return res(nums,k,sum/k,0,0,visited); } bool res(vector<int> &nums, int k, int target,int start,int cursum, vector<bool>& visited) { if (k == 1) return true; if (cursum == target) return res(nums,k-1,target,0,0,visited); for (int i = start; i< nums.size(); i++) { if (visited[i]) continue; visited[i] = true; if (res(nums,k,target,i+1,cursum+nums[i],visited)) return true; visited[i] = false; } return false; }};
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