Give you two non-negative integer arrays rowSum
and colSum
, where rowSum[i]
is the in the two-dimensional matrix. /span> column elements. In other words, you don’t know every element in the matrix, but you know the sum of each row and column. is the sum of the i
The sum of the row elements, colSum[j]
j
Please find any non-negative integer matrix of size rowSum.length x colSum.length
, and the matrix satisfies requirements. and rowSum
colSum
Please return any two-dimensional matrix that meets the requirements of the question. The question is guaranteed to exist at least one feasible matrix.
Example 1:
Input:rowSum = [3,8], colSum = [4,7] Output: [[3,0], [1,7]]Explanation: Row 0: 3 + 0 = 3 == rowSum[0] Row 1: 1 + 7 = 8 == rowSum[1] Column 0: 3 + 1 = 4 == colSum[0] Column 1: 0 + 7 = 7 == colSum[1] The sums of rows and columns meet the question requirements, and all matrix elements are non-negative. Another feasible matrix is: [[1,2], [3,5]]
Example 2:
Import:rowSum = [5,7,10], colSum = [8,6,8] Output:[[0,5,0], [6,1,0], [2,0,8]]
Example 3:
Import:rowSum = [14,9], colSum = [6,9,8] Output:[[0,9,5], [6,0,3]]
Example 4:
Import:rowSum = [1,0], colSum = [1] Output:[[1], [0]]
Example 5:
Import:rowSum = [0], colSum = [0] Output:[[0]]
hint:
1 <= rowSum.length, colSum.length <= 500
0 <= rowSum[i], colSum[i] <= 108
sum(rowSum) == sum(colSum)
思路:
Starting from the upper left corner, remove either a row or a column at a time. Every time min is found, just subtract it according to the conditions to determine whether to wrap the line or the column.
class Solution {
public:
vector<vector<int>> restoreMatrix(vector<int>& rowSum, vector<int>& colSum) {//贪心
int row = rowSum.size();
int col = colSum.size();
vector<vector<int>> ans(row,vector<int>(col));//初始化二维数组
for(int i=0,j=0;i<row&&j<col;){
if(rowSum[i]>colSum[j]){//换列
rowSum[i]-=colSum[j];
ans[i][j++] = colSum[j];
}
else {//换行
colSum[j]-=rowSum[i];
ans[i++][j] = rowSum[i];
}
}
return ans;
}
};