给你两个非负整数数组 rowSum
和 colSum
,其中 rowSum[i]
是二维矩阵中第 i
行元素的和, colSum[j]
是第 j
列元素的和。换言之你不知道矩阵里的每个元素,但是你知道每一行和每一列的和。
请找到大小为 rowSum.length x colSum.length
的任意 非负整数 矩阵,且该矩阵满足 rowSum
和 colSum
的要求。
请你返回任意一个满足题目要求的二维矩阵,题目保证存在 至少一个 可行矩阵。
示例 1:
输入:rowSum = [3,8], colSum = [4,7] 输出:[[3,0], [1,7]] 解释: 第 0 行:3 + 0 = 3 == rowSum[0] 第 1 行:1 + 7 = 8 == rowSum[1] 第 0 列:3 + 1 = 4 == colSum[0] 第 1 列:0 + 7 = 7 == colSum[1] 行和列的和都满足题目要求,且所有矩阵元素都是非负的。 另一个可行的矩阵为:[[1,2], [3,5]]
示例 2:
输入:rowSum = [5,7,10], colSum = [8,6,8] 输出:[[0,5,0], [6,1,0], [2,0,8]]
示例 3:
输入:rowSum = [14,9], colSum = [6,9,8] 输出:[[0,9,5], [6,0,3]]
示例 4:
输入:rowSum = [1,0], colSum = [1] 输出:[[1], [0]]
示例 5:
输入:rowSum = [0], colSum = [0] 输出:[[0]]
提示:
1 <= rowSum.length, colSum.length <= 500
0 <= rowSum[i], colSum[i] <= 108
sum(rows) == sum(columns)
class Solution:
def restoreMatrix(self, rowSum: List[int], colSum: List[int]) -> List[List[int]]:
m, n = len(rowSum), len(colSum)
ans = [[0] * n for _ in range(m)]
for i in range(m):
for j in range(n):
x = min(rowSum[i], colSum[j])
ans[i][j] = x
rowSum[i] -= x
colSum[j] -= x
return ans
class Solution {
public int[][] restoreMatrix(int[] rowSum, int[] colSum) {
int m = rowSum.length;
int n = colSum.length;
int[][] ans = new int[m][n];
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
int x = Math.min(rowSum[i], colSum[j]);
ans[i][j] = x;
rowSum[i] -= x;
colSum[j] -= x;
}
}
return ans;
}
}
class Solution {
public:
vector<vector<int>> restoreMatrix(vector<int>& rowSum, vector<int>& colSum) {
int m = rowSum.size(), n = colSum.size();
vector<vector<int>> ans(m, vector<int>(n));
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
int x = min(rowSum[i], colSum[j]);
ans[i][j] = x;
rowSum[i] -= x;
colSum[j] -= x;
}
}
return ans;
}
};
func restoreMatrix(rowSum []int, colSum []int) [][]int {
m, n := len(rowSum), len(colSum)
ans := make([][]int, m)
for i := range ans {
ans[i] = make([]int, n)
}
for i := range rowSum {
for j := range colSum {
x := min(rowSum[i], colSum[j])
ans[i][j] = x
rowSum[i] -= x
colSum[j] -= x
}
}
return ans
}
func min(a, b int) int {
if a < b {
return a
}
return b
}