标签：
题目链接：https://leetcode.com/problems/range-sum-query-2d-immutable/
题目：
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).
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:
思路：
简单。。。
算法：
int s[][];
public NumMatrix(int[][] matrix) {
if (matrix.length != 0) {
s = new int[matrix.length][matrix[0].length];
s[0][0] = matrix[0][0];
for (int i = 1; i < matrix.length; i++) {
s[i][0] = s[i - 1][0] + matrix[i][0];
}
for (int j = 1; j < matrix[0].length; j++) {
s[0][j] = s[0][j - 1] + matrix[0][j];
}
for (int i = 1; i < matrix.length; i++) {
for (int j = 1; j < matrix[0].length; j++) {
s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + matrix[i][j];
}
}
}
}
public int sumRegion(int row1, int col1, int row2, int col2) {
int s2 = 0, s3 = 0, s4 = 0;
if (col1 - 1 >= 0)
s2 = s[row2][col1 - 1];
if (row1 - 1 >= 0)
s3 = s[row1 - 1][col2];
if (row1 - 1 >= 0 && col1 - 1 >= 0)
s4 = s[row1 - 1][col1 - 1];
return s[row2][col2] - s2 - s3 + s4;
} 【Leetcode】Range Sum Query 2D - Immutable
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原文地址：http://blog.csdn.net/yeqiuzs/article/details/51542400
