作者:轰炸籹厕所744 | 来源:互联网 | 2023-10-13 16:56
Givena2Dmatrixmatrix,findthesumoftheelementsinsidetherectangledefinedbyitsupperleftcorner(
Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).
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The above rectangle (with the red border) is defined by (row1, col1) = (2, 1) and (row2, col2) = (4, 3), which contains sum = 8.
Example:
Given matrix = [
[3, 0, 1, 4, 2],
[5, 6, 3, 2, 1],
[1, 2, 0, 1, 5],
[4, 1, 0, 1, 7],
[1, 0, 3, 0, 5]
]
sumRegion(2, 1, 4, 3) -> 8
sumRegion(1, 1, 2, 2) -> 11
sumRegion(1, 2, 2, 4) -> 12
Note:
- You may assume that the matrix does not change.
- There are many calls to sumRegion function.
- You may assume that row1 ≤ row2 and col1 ≤ col2.
题目大意:
求解一个矩阵范围内的数字和。
解法:
构造一个二维数组和[sum+1][col+1],sum[i+1][j+1]表示从matrix[0][0]到matrix[i][j]的数字之和。
class NumMatrix {
private int m,n;
int[][] sum;
public NumMatrix(int[][] matrix) {
m=matrix.length;
n=(m==0?0:matrix[0].length);
sum=new int[m+1][n+1];
for (int i=1;i<=m;i++){
for (int j=1;j<=n;j++){
sum[i][j]=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]+matrix[i-1][j-1];
}
}
}
public int sumRegion(int row1, int col1, int row2, int col2) {
return sum[row2+1][col2+1]-sum[row1][col2+1]-sum[row2+1][col1]+sum[row1][col1];
}
}