Partition to K Equal Sum Subsets
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题目描述
解题思路
题目大意如下,给你一个一维int数组,且每个数大于0,再给你一个数字k,问是否可以将该数组分成k个子集,这k个子集的并集刚好为该int数组,交集为空,最关键的是,这k个子集每个子集的数字之和都相等。
仔细思考不难发现,要想满足题意,首先应该判断这个int数组的总和sum是否可以整除k,若能整除则就到了本题目的难点,怎么才能刚好划分成k个子集每个子集的数字和等于sum/k呢?
我想到的是先对数组进行排序,之后从大到小开始递推考虑,假设函数gofind(int target, int index)代表从下标index开始,能否从数组中找到某几个数,他们的和刚好等于target呢?(并且我优先考虑大数,相加的这些数令其为0以表示被用过下次不能再被使用,即从大数组中删掉。)那么gofind(int target, int index)便取决于gofind(target-nums[i], i-1)啦。
话不多说,上代码~
C++代码实现
class Solution {public: bool gofind(vector<int>& nums, int target, int index, int& countt) { //cout << "要在前" << index+1 << "个数字找有没有和为" << target << "的" << endl; if (target < 0) { return false; } // 能否以当前下标index开始,寻找某几个数字,加起来和为target //(优先考虑大数,相加的这些数令其为0以表示被用过) for (int i = index; i >= 0; --i) { if (nums[i] == 0 || nums[i] > target) { continue; } else if (nums[i] == target) { countt++; //cout << nums[i] << " " << i << endl; nums[i] = 0; return true; } else { //cout << target << "-(" << i << ")" << nums[i] << endl; if (gofind(nums, target-nums[i], i-1, countt)) { nums[i] = 0; return true; } } } return false;} bool canPartitionKSubsets(vector<int>& nums, int k) { if (k == 1) { return true; } int sum = 0; int length = nums.size(); for (int i = 0; i < length; ++i) { sum += nums[i]; } if (sum % k != 0) { return false; } int target = sum / k; int countt = 0; //cout << sum << " " << target << endl; sort(nums.begin(), nums.end()); for (int i = length - 1; i >= 0; --i) { if (nums[i] > target) { return false; } else if (nums[i] == target) { countt++; //cout << countt << " " << nums[i] << " " << i << endl; nums[i] = 0; // 已被标记 } else if (nums[i] == 0) { continue; } else { if (!gofind(nums, target, i, countt)) { return false; } } } for (int i = 0; i < length; ++i) { if (nums[i] != 0) { // cout << nums[i] << " " << i << endl; return false; } } //cout << countt << endl; return countt == k; }};
题目样例运行截图
反思
感觉这道题的思路还算ok,不过复杂度的话确实蛮高的。而且在实现过程中,没能很好的考虑函数传参的问题,导致debug了一晚。。。总的来说,收获蛮多的~
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