@Yano
2016-03-21T20:34:49.000000Z
字数 1406
阅读 2875
LeetCode
Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (_row_1, _col_1) and lower right corner (_row_2, _col_2).
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:
和 303 题一样,只不过变成二维数组。那么就分别把每一行用 303 的解法。
public class NumMatrix {
public long[][] sumMatrix;
public NumMatrix(int[][] matrix) {
if (matrix == null || matrix.length == 0) {
return;
}
sumMatrix = new long[matrix.length + 1][matrix[0].length + 1];
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix[0].length; j++) {
sumMatrix[i][j + 1] = sumMatrix[i][j] + matrix[i][j];
}
}
}
public int sumRegion(int row1, int col1, int row2, int col2) {
if (sumMatrix == null || row1 < 0 || row2 < 0 || col1 < 0
|| col2 < 0 || row1 >= sumMatrix.length - 1
|| row2 >= sumMatrix.length - 1
|| col1 >= sumMatrix[0].length - 1
|| col2 >= sumMatrix[0].length - 1 || row1 > row2
|| col1 > col2) {
return Integer.MIN_VALUE;
}
long rt = 0;
for (int i = row1; i <= row2; i++) {
rt += sumMatrix[i][col2 + 1] - sumMatrix[i][col1];
}
return (int) rt;
}
}
// Your NumMatrix object will be instantiated and called as such:
// NumMatrix numMatrix = new NumMatrix(matrix);
// numMatrix.sumRegion(0, 1, 2, 3);
// numMatrix.sumRegion(1, 2, 3, 4);