Leetcode | Partition to K Equal Sum Subsets
2017-11-12 23:02
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原题链接:https://leetcode.com/problems/partition-to-k-equal-sum-subsets
原题意如下所示:
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:
Note:
1 <= k <= len(nums) <= 16.
0 < nums[i] < 10000.
题目描述的意思大概就是给定一个数组nums,然后给定一个整数k,问的是能不能把数组nums划分成k部分,而且这k部分中每一部分的和还要相等,这里要注意的是是将整个数组进行划分,所以数组中的每一个数字有且只能出现在其中一个部分中,这样的话这道题就不会显得那么复杂了。
我们由上述描述可以从整个数组中提前得出每一部分的和应为nums总和sum/k,所以由此我们可以想到这类似背包问题,即将一个部分看成一个背包,要从数组中选取一些数字让它们恰好等于sum/k的值,用递归的方式解决就必须有终止条件,在这个问题中,最后的临界条件即是只剩一个背包时,肯定返回的是true,还有另外一个临界条件,即当一个背包里的值恰好为sum/k时,那么就可以继续搜索剩下的k-1个背包,而当背包里的总和>sum/k或将数组里的数都装满时,就返回false。
源代码如下所示:
原题意如下所示:
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:
Note:
1 <= k <= len(nums) <= 16.
0 < nums[i] < 10000.
题目描述的意思大概就是给定一个数组nums,然后给定一个整数k,问的是能不能把数组nums划分成k部分,而且这k部分中每一部分的和还要相等,这里要注意的是是将整个数组进行划分,所以数组中的每一个数字有且只能出现在其中一个部分中,这样的话这道题就不会显得那么复杂了。
我们由上述描述可以从整个数组中提前得出每一部分的和应为nums总和sum/k,所以由此我们可以想到这类似背包问题,即将一个部分看成一个背包,要从数组中选取一些数字让它们恰好等于sum/k的值,用递归的方式解决就必须有终止条件,在这个问题中,最后的临界条件即是只剩一个背包时,肯定返回的是true,还有另外一个临界条件,即当一个背包里的值恰好为sum/k时,那么就可以继续搜索剩下的k-1个背包,而当背包里的总和>sum/k或将数组里的数都装满时,就返回false。
源代码如下所示:
class Solution { public: bool canPartitionKSubsets(vector<int>& nums, int k) { int sum = 0; for(int num:nums)sum+=num; if(k <= 0 || sum%k != 0)return false; vector<int> visited(nums.size(), 0); return canPartition(nums, visited, 0, k, 0, 0, sum/k); } bool canPartition(vector<int>& nums, vector<int>& visited, int start_index, int k, int cur_sum, int cur_num, int target){ if(k==1)return true; if(cur_sum > target) return false; if(cur_sum == target && cur_num >0 )return canPartition(nums, visited, 0, k-1, 0, 0, target); for(int i = start_index; i<nums.size(); i++){ if(!visited[i]){ visited[i] = 1; if(canPartition(nums, visited, i+1, k, cur_sum + nums[i], cur_num++, target))return true; visited[i] = 0; } } return false; } };
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