leetcode 1605. Find Valid Matrix Given Row and Column Sums（python）

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描述

You are given two arrays rowSum and colSum of non-negative integers where rowSum[i] is the sum of the elements in the i^th row and colSum[j] is the sum of the elements of the j^th column of a 2D matrix. In other words, you do not know the elements of the matrix, but you do know the sums of each row and column.

Find any matrix of non-negative integers of size rowSum.length x colSum.length that satisfies the rowSum and colSum requirements.

Return a 2D array representing any matrix that fulfills the requirements. It's guaranteed that at least one matrix that fulfills the requirements exists.

Example 1:

Input: rowSum = [3,8], colSum = [4,7]
Output: [[3,0],
         [1,7]]
Explanation:
0th row: 3 + 0 = 3 == rowSum[0]
1st row: 1 + 7 = 8 == rowSum[1]
0th column: 3 + 1 = 4 == colSum[0]
1st column: 0 + 7 = 7 == colSum[1]
The row and column sums match, and all matrix elements are non-negative.
Another possible matrix is: [[1,2],
                             [3,5]]	
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Example 2:

Input: rowSum = [5,7,10], colSum = [8,6,8]
Output: [[0,5,0],
         [6,1,0],
         [2,0,8]]
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Example 3:

Input: rowSum = [14,9], colSum = [6,9,8]
Output: [[0,9,5],
         [6,0,3]]
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Example 4:

Input: rowSum = [1,0], colSum = [1]
Output: [[1],
         [0]]
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Example 5:

Input: rowSum = [0], colSum = [0]
Output: [[0]]
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Note:

1 <= rowSum.length, colSum.length <= 500
0 <= rowSum[i], colSum[i] <= 10^8
sum(rows) == sum(columns)
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解析

根据题意，就是有一个矩阵，但是没有给出具体每个位置的值，而是给出来每个行的总和列表 rowSum ，每个列的总和列表 colSum ，让我们推断出一个矩阵，让其满足对应的行、列的总和 rowSum 、colSum 的要求，这种矩阵可以有多个，给出一个满足题意的结果就可以了。思路比较简单，用贪心算法：

• 先初始化一个全 0 的矩阵 result ，行数为 rowSum 的个数，列数为 colSum 的个数

• 遍历 result 中的每个位置，因为每个位置的数值大小肯定不大于 min(当前行总和, 当前列总和)，所以可以先暂时设置

    result[i][j] = min(rowSum[i], colSum[j]
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• 然后将当前行和当前列的总和都减去 result[i][j] ，继续进行下一轮的数值计算

• 遍历结束得到的 result 即为结果

解答

class Solution(object):
    def restoreMatrix(self, rowSum, colSum):
        """
        :type rowSum: List[int]
        :type colSum: List[int]
        :rtype: List[List[int]]
        """
        M = len(rowSum)
        N = len(colSum)
        result = [[0]*N for _ in range(M)]
        for i in range(M):
            for j in range(N):
                result[i][j] = min(rowSum[i], colSum[j])
                rowSum[i] -= result[i][j]
                colSum[j] -= result[i][j]
        return result
        	      
		
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运行结果

Runtime: 888 ms, faster than 76.00% of Python online submissions for Find Valid Matrix Given Row and Column Sums.
Memory Usage: 18.2 MB, less than 66.00% of Python online submissions for Find Valid Matrix Given Row and Column Sums.
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原题链接：leetcode.com/problems/fi…

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