698[Medium]: Partition to K Equal Sum Subsets

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本文探讨了一道关于将数组等分为多个子集的问题，并提供了一种解决方案。输入为一整数数组与目标子集数量，目标是判断是否能将数组等分为指定数量的子集，且各子集之和相等。文章通过示例说明了问题背景，并分享了一个基于回溯搜索的解题思路及代码实现。

Part1:题目描述

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.

Part2:解题思路

这一题我觉得对我来说还是挺难的，借鉴了leetcode上面的解法才得以解决，要再更深入的理解才行

Part3:代码

class Solution {
public:
    bool search(vector<int>groups, int row, vector<int>& nums, int target) {
	
	if (row < 0) return true;
	// 执行完这一句之后row的值才会变 
	int v = nums[row--];
	for (int i = 0; i < groups.size(); i++) {
		if (groups[i] + v <= target) {
			groups[i] += v;
			if (search(groups, row, nums, target)) return true;
			groups[i] -= v;
		}
		if (groups[i] == 0) break;
	}
	return false;
}

bool canPartitionKSubsets(vector<int>& nums, int k) {
	int length = nums.size();
	int sum = 0;
	for (int i = 0; i < length; i++) {
		sum += nums[i];
	}
	if ((sum % k) != 0) return false;
	int target = sum / k;
	sort(nums.begin(), nums.end());
	
	
	
	int row = length - 1;
	if(nums[row] > target) return false;
	while (row >= 0 && nums[row] == target) {
		row--;
		k--;
	} 
	return search(vector<int>(k), row, nums, target);
}
};