@Yano 2016-03-21T12:34:49.000000Z 字数 1406 阅读 2817

# LeetCode 304 Range Sum Query 2D - Immutable

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:

1. You may assume that the matrix does not change.
2. There are many calls to sumRegion function.
3. You may assume that _row_1 ≤ _row_2 and _col_1 ≤ _col_2.

# 代码

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);

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