原题链接在这里：https://leetcode.com/problems/partition-to-k-equal-sum-subsets/description/
题目：
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.
题解：
首先计算sum, 看sum能否被k整除. 若不能, 铁定不能分成k组. return false.
若能的话，每组的target sum就该是sum/k. 一组一组的减掉. 直到 k = 1. 剩下最后一组, 最后一组的sum肯定是sum/k.
因为这里的已经验证过sum是k的倍数, 而前面已经有k-1组 sum/k找到了. 所以可以直接return true.
This is bottom-up recursion. Set parameters for state first.
It needs count to count number in subarray. Since there may be negative number in nums. If target is 0, there could be [-1, 1] or empty subarray.
The reason state has both visited and cur starting index is because of trimming dfs tree.
When summing up to target, if index i can't be used, when trying j > i, the next level of DFS, there is no need to try i again. Because if i works, it would be added into res before.
The only case i could be used is to sum up next target.
Note: the question is asking for non-empty, we need to add a count of each sub set. And make sure it is > 0 before accumlating to result.
Time Complexity: exponential.
Space: O(n). stack space.
AC Java:
class Solution {
public boolean canPartitionKSubsets(int[] nums, int k) {
if(nums == null || nums.length == 0){
return false;
}
int sum = 0;
for(int num : nums){
sum += num;
}
if(sum % k != 0){
return false;
}
boolean [] visited = new boolean[nums.length];
return dfs(nums, visited, 0, 0, sum/k, 0, k);
}
private boolean dfs(int [] nums, boolean [] visited, int cur, int sum, int target, int count, int k){
if(sum > target){
return false;
}
if(k == 1){
return true;
}
if(sum == target && count > 0){
return dfs(nums, visited, 0, 0, target, 0, k-1);
}
for(int i = cur; i<nums.length; i++){
if(!visited[i]){
visited[i] = true;
if(dfs(nums, visited, i+1, sum+nums[i], target, count++, k)){
return true;
}
visited[i] = false;
}
}
return false;
}
}