QUESTION:
You are given two arrays rowSum and colSum of non-negative integers where rowSum[i] is the sum of the elements in the ith row and colSum[j] is the sum of the elements of the jth 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]]
Example 2:
Input: rowSum = [5,7,10], colSum = [8,6,8]
Output: [[0,5,0],
[6,1,0],
[2,0,8]]
Constraints:
1 <= rowSum.length, colSum.length <= 500
0 <= rowSum[i], colSum[i] <= 108
sum(rows) == sum(columns)
EXPLANATION:
用贪心算法, 在每一个位置放置row和colum的最小值, 同时记录下当前剩下的值即可.
SOLUTION:
class Solution {
func restoreMatrix(_ rowSum: [Int], _ colSum: [Int]) -> [[Int]] {
let rc = rowSum.count, cc = colSum.count
var rowSum = rowSum, colSum = colSum, result = [[Int]](repeating: [Int](repeating: 0, count: cc), count: rc)
for r in 0..<rc {
for c in 0..<cc {
result[r][c] = min(rowSum[r], colSum[c])
rowSum[r] -= result[r][c]
colSum[c] -= result[r][c]
}
}
return result
}
}