Description
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]]
Example 3:
Input: rowSum = [14,9], colSum = [6,9,8]
Output: [[0,9,5],
[6,0,3]]
Example 4:
Input: rowSum = [1,0], colSum = [1]
Output: [[1],
[0]]
Example 5:
Input: rowSum = [0], colSum = [0]
Output: [[0]]
Constraints:
- 1 <= rowSum.length, colSum.length <= 500
- 0 <= rowSum[i], colSum[i] <= 108
- sum(rows) == sum(columns)
分析
题目的意思是：给定rowSum,表示每一行的和，colSum，表示每一列的和。现在根据rowSum和colSum来重构出一个矩阵。
- 参考了一下答案，用的贪心法，遍历rowSum，和colSum数组，首先填上rowSum和colSum中最小的值，如果最小值是rowSum最小，填上后，colSum进行更新，rowSum遍历位置+1；如果最小值是colSum，填上后，rowSum进行更新，colSum遍历的位置+1
这个解法就厉害了，至少我想不出来。我验证了一下，确实是正确的。
代码
class Solution:
def restoreMatrix(self, rowSum: List[int], colSum: List[int]) -> List[List[int]]:
m=len(rowSum)
n=len(colSum)
res=[[0]*n for i in range(m)]
i=0
j=0
while(i<m and j<n):
res[i][j]=min(rowSum[i],colSum[j])
if(rowSum[i]==colSum[j]):
i+=1
j+=1
elif(rowSum[i]>colSum[j]):
rowSum[i]-=colSum[j]
j+=1
else:
colSum[j]-=rowSum[i]
i+=1
return res
参考文献
Python Beats 99.83% with Illustration
