Description：

• Difficulty:Medium
• Total Accepted:7.1K
• Total Submissions:19.3K
• Contributor: fishercoder

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.

Thought：

本题虽然在动态规划的范围内，但是并没有发现特别好的利用动态规划的解法。

如果sum/k不能整除，或者nums中存在任意比k大的数，则无解。

接着nums中的每个元素对应k个状态，所有有nk中情况，dfs用到了剪枝，排序贪心尽早把不合法的解从递归树中删除。

Code：

class Solution {
public:
	bool canPartitionKSubsets(vector<int>& nums, int k) {
		int max = 0;
		int sum = 0;
		for (int i = 0; i < nums.size(); i++) {
			sum += nums[i];
			max = max < nums[i] ? nums[i] : max;
		}
		if (sum%k != 0)return false;
		if (max > sum / k)return false;
		
        	vector<int> memo(k, 0);
        	return search(nums, 0,nums.size(),k,sum/k,memo);
	}

	bool search(vector<int>& nums, int pos, int n, int k, int target, vector<int>& memo) {
		if (pos == n) return true; // n,
		int jEnd = std::min(k, pos + 1);
		for (int j = 0; j < jEnd; ++j) {
			if (nums[pos] + memo[j] > target) {
				continue;
			}
			memo[j] += nums[pos];
			if (search(nums, pos + 1, n, k, target, memo)) {
				return true;
			}
			memo[j] -= nums[pos];
		}
		return false;
	}
};