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n^k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;
}