698.Partition to K Equal Sum Subsets
curNum {How many num have been used, incase of sum = -}
public boolean canPartitionKSubsets(int[] nums, int k) {
int sum = 0;
for(int num:nums)sum += num;
if(k <= 0 || sum%k != 0)return false;
int[] visited = new int[nums.length];
return canPartition(nums, visited, 0, k, 0, 0, sum/k);
}
public boolean canPartition(int[] nums, int[] visited, int start_index, int k, int cur_sum, int cur_num, int target){
if(k==1)return true;
if(cur_sum == target && cur_num>0)return canPartition(nums, visited, 0, k-1, 0, 0, target);
for(int i = start_index; i<nums.length; i++){
if(visited[i] == 0){
visited[i] = 1;
if(canPartition(nums, visited, i+1, k, cur_sum + nums[i], cur_num++, target))return true;
visited[i] = 0;
}
}
return false;
}