CSDN Topic Challenge Phase 2
Participation Topic: Algorithm Solution
Article directory
Topic link and description
https://leetcode.cn/problems/partition-to-k-equal-sum-subsets/
Given an integer array nums and a positive integer k, find whether it is possible to divide this array into k non-empty subsets whose sums are equal.
Example 1:
Input: nums = [4, 3, 2, 3, 5, 2, 1], k = 4
Output: True
Explanation: It is possible to split it into 4 subsets (5), (1,4), (2,3 ), (2,3) is equal to the sum.
Example 2:
Input: nums = [1,2,3,4], k = 3
Output: false
hint:
1 <= k <= len(nums) <= 16
0 < nums[i] < 10000
The frequency of each element is in the range [1,4]
Keywords: recursion
method one:
run screenshot
the code
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The array is divided into k non-empty subsets and the sum is equal. The sum can be known as: sum(nums)/k = sum of subsets m
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If the remainder here is not 0, it means that it cannot be divided equally
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At the same time, the elements are greater than 0, and the maximum value is greater than m, which cannot be combined
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It is conceivable that after sorting first, the double pointers are summed, and the set is assembled to make it equal to m
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But the left pointer needs to match the big ones as much as possible, because the small ones can be more than a few to form the average value, and the large numbers cannot be grouped, and the left pointer starts from the right pointer, so use a serial number to recursively traverse
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At the beginning, the variable is stored in advance, and recursion is required later to avoid the trouble of passing parameters
int[] numsField;
int index, avg, kNum;
public boolean canPartitionKSubsets(int[] nums, int k) {
numsField = nums;
kNum = k;
int total = 0;
// 计算总值
for (int x : numsField) {
total += x;
}
// 平均值不等显然无法凑齐
if (total % kNum != 0) {
return false;
}
// 排序后遍历
Arrays.sort(numsField);
index = numsField.length; avg = total / kNum;
return dfs(index - 1, 0, 0, new boolean[index]);
}
boolean dfs(int idx, int currentSum, int countK, boolean[] vis) {
// 如果统计的k数等于预期的，完成
if (countK == kNum) {
return true;
}
// 如果当前累积已经满足平均值，接着遍历下一个数
if (currentSum == avg) {
return dfs(index - 1, 0, countK + 1, vis);
}
// 如果遍历完成，却cur不等avg
if (idx == -1) {
return false;
}
// 继续迭代，遍历查询
for (int i = idx; i >= 0; i--) {
// 访问过 或者 大于 显然不符合
if (vis[i] || currentSum + numsField[i] > avg) {
continue;
}
// 小于或者等于 则设为访问过
vis[i] = true;
// 将下一个推入，如果是符合的则返回true
if (dfs(i - 1, currentSum + numsField[i], countK, vis)) {
return true;
}
// 不符合要将恢复false ， 可能存在后续的数组使用
vis[i] = false;
// 当前累积值如果是0 直接返回false 说明已经没有数了
if (currentSum == 0) {
return false;
}
}
// 最后都没有找到
return false;
}
end
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