标签:相加 inpu leetcode problem for 二维 order discus put
You are given an m * n
matrix, mat
, and an integer k
, which has its rows sorted in non-decreasing order.
You are allowed to choose exactly 1 element from each row to form an array. Return the Kth smallest array sum among all possible arrays.
Example 1:
Input: mat = [[1,3,11],[2,4,6]], k = 5 Output: 7 Explanation: Choosing one element from each row, the first k smallest sum are: [1,2], [1,4], [3,2], [3,4], [1,6]. Where the 5th sum is 7.
Example 2:
Input: mat = [[1,3,11],[2,4,6]], k = 9 Output: 17
Example 3:
Input: mat = [[1,10,10],[1,4,5],[2,3,6]], k = 7 Output: 9 Explanation: Choosing one element from each row, the first k smallest sum are: [1,1,2], [1,1,3], [1,4,2], [1,4,3], [1,1,6], [1,5,2], [1,5,3]. Where the 7th sum is 9.
Example 4:
Input: mat = [[1,1,10],[2,2,9]], k = 7 Output: 12
Constraints:
m == mat.length
n == mat.length[i]
1 <= m, n <= 40
1 <= k <= min(200, n ^ m)
1 <= mat[i][j] <= 5000
mat[i]
is a non decreasing array.
题意:
给出一个二维数组,数组的每一行按照非减的顺序进行排列,在每一行中选出一个数字,求出所选数字的和,将这些数字排序,并求出第k小的那个。
思路:
本来想的是用DFS进行求解,如果这样做的话,最欢情况下的时间复杂度会是O(40^40),用DFS遍历求解的话,有可能会出现后面出现的数字比前面出现数字要小的情况,好像没办法剪枝,这种方法行不通。换种思路,我们可以用相邻两行进行相加,求得的和进行排序,截取前面的k个数字。然后再将所求的和与下一行元素相加。
Code:
1 class Solution { 2 public: 3 int kthSmallest(vector<vector<int>>& mat, int k) { 4 vector<int> ans = {0}, temp; 5 for (int i = 0; i < mat.size(); ++i) { 6 for (int j = 0; j < mat[i].size(); ++j) { 7 for (int v = 0; v < ans.size(); ++v) { 8 temp.push_back(mat[i][j] + ans[v]); 9 } 10 } 11 sort(temp.begin(), temp.end()); 12 ans.clear(); 13 int len = min(k, (int)temp.size()); 14 for (int j = 0; j < len; ++j) { 15 ans.push_back(temp[j]); 16 } 17 temp.clear(); 18 } 19 return ans[k-1]; 20 } 21 };
参考:
5403. Find the Kth Smallest Sum of a Matrix With Sorted Rows
标签:相加 inpu leetcode problem for 二维 order discus put
原文地址:https://www.cnblogs.com/ruruozhenhao/p/12822640.html