LeetCode（304）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).

Range Sum Query 2D
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.

分析

思想与上一题相同。

注意下标处理方式。

AC代码

class NumMatrix {public:    NumMatrix(vector<vector<int>> &matrix) {        if (matrix.empty())            return;        //求得二维矩阵的行列        int m = matrix.size(), n = matrix[0].size();        sums = vector<vector<int>>(m+1, vector<int>(n+1, 0));        sums[0][0] = matrix[0][0];        for (int j = 1; j <= n; ++j)        {            sums[0][j] = 0;        }        for (int i = 1; i <= m; ++i)        {            sums[i][0] = 0;        }        //        for (int i = 1; i <= m; ++i)        {            for (int j = 1; j <= n; ++j)            {                sums[i][j] = sums[i][j - 1] + sums[i - 1][j] - sums[i - 1][j - 1] + matrix[i-1][j-1];            }//for        }//for          }    int sumRegion(int row1, int col1, int row2, int col2) {        //求得二维矩阵的行列        int m = sums.size(), n = sums[0].size();        if (row1 < 0 || row1 >= m || col1 < 0 || col1 >= n || row2<0 || row2 >= m ||            col2<0 || col2 >= n || row1 >row2 || col1 > col2)        {            return 0;        }        return sums[row2+1][col2+1] - sums[row2+1][col1] - sums[row1][col2+1] + sums[row1][col1];    }private:    vector<vector<int>> sums;};

GitHub测试程序源码