LWC 54：698. Partition to K Equal Sum Subsets

LWC 54：698. Partition to K Equal Sum Subsets

传送门：698. Partition to K Equal Sum Subsets

Problem:

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.

思路：

观察 k 和 n 发现均很小，所以实际上是暴力dfs算法，先预处理，如果sum / k 有余数，则不能分割。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。

代码如下：

public boolean canPartitionKSubsets(int[] nums, int k) {
int sum = 0;
int max = 0;
int n = nums.length;
for (int i = 0; i < n; ++i) {
sum += nums[i];
max = Math.max(max, nums[i]);
}
if (sum % k != 0) return false;
tar = sum / k;
if (max > tar) return false;
Arrays.sort(nums);
return go(nums, n - 1, k, new int[k]);
}

int tar = 0;
boolean go(int[] nums, int pos, int k, int[] sums) {
if (pos == -1) {
boolean check = true;
for (int i = 0; i < k; ++i) {
if (sums[i] != tar) check = false;
}
return check;
}
for (int i = 0; i < k; ++i) {
sums[i] += nums[pos];
if (sums[i] <= tar && go(nums, pos - 1, k, sums)) {
return true;
}
sums[i] -= nums[pos];
}
return false;
}