1. Description
2. Solution
**解析:**Version 1,貪心算法,由于矩陣中的每一個元素
matrix[i][j]
一定不大于
min(rowSum[i], colSum[j]
,是以将
matrix[i][j]
設定為
min(rowSum[i], colSum[j]
就可以解決一行或一列的數值設定問題,目前行或目前列剩餘的元素為
,周遊所有元素存在重複設定的問題。Version 2進行了優化,減少了重複設定的問題,如果輸出Version 1最終的
rowSum, colSum
,會發現所有的元素都變為了0,而Version 2的則不是,Version對于不再用到的
rowSum[i], colSum[j]
,沒有減去
matrix[i][j]
。
- Version 1
class Solution:
def restoreMatrix(self, rowSum: List[int], colSum: List[int]) -> List[List[int]]:
m = len(rowSum)
n = len(colSum)
matrix = [n * [0] for _ in range(m)]
for i in range(m):
for j in range(n):
matrix[i][j] = min(rowSum[i], colSum[j])
rowSum[i] -= matrix[i][j]
colSum[j] -= matrix[i][j]
return matrix
複制
- Version 2
class Solution:
def restoreMatrix(self, rowSum: List[int], colSum: List[int]) -> List[List[int]]:
m = len(rowSum)
n = len(colSum)
matrix = [n * [0] for _ in range(m)]
i = 0
j = 0
while i < m and j < n:
if rowSum[i] > colSum[j]:
matrix[i][j] = colSum[j]
rowSum[i] -= colSum[j]
j += 1
elif rowSum[i] < colSum[j]:
matrix[i][j] = rowSum[i]
colSum[j] -= rowSum[i]
i += 1
else:
matrix[i][j] = rowSum[i]
i += 1
j += 1
return matrix
複制
Reference
- https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/