leetcode 304. Range Sum Query 2D

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) -> 8sumRegion(1, 1, 2, 2) -> 11sumRegion(1, 2, 2, 4) -> 12

Note:

1. You may assume that the matrix does not change.
2. There are many calls to sumRegion function.

1. You may assume that row1 ≤ row2 and col1 ≤ col2.

解法：

+-----+-+-------+     +--------+-----+     +-----+---------+     +-----+--------+|     | |       |     |        |     |     |     |         |     |     |        ||     | |       |     |        |     |     |     |         |     |     |        |+-----+-+       |     +--------+     |     |     |         |     +-----+        ||     | |       |  =  |              |  +  |     |         |  -  |              |+-----+-+       |     |              |     +-----+         |     |              ||               |     |              |     |               |     |              ||               |     |              |     |               |     |              |+---------------+     +--------------+     +---------------+     +--------------+   sums[i][j]      =    sums[i-1][j]    +     sums[i][j-1]    -   sums[i-1][j-1]   +                          matrix[i-1][j-1]

So, we use the same idea to find the specific area's sum.

+---------------+   +--------------+   +---------------+   +--------------+   +--------------+|               |   |         |    |   |   |           |   |         |    |   |   |          ||   (r1,c1)     |   |         |    |   |   |           |   |         |    |   |   |          ||   +------+    |   |         |    |   |   |           |   +---------+    |   +---+          ||   |      |    | = |         |    | - |   |           | - |      (r1,c2) | + |   (r1,c1)    ||   |      |    |   |         |    |   |   |           |   |              |   |              ||   +------+    |   +---------+    |   +---+           |   |              |   |              ||        (r2,c2)|   |       (r2,c2)|   |   (r2,c1)     |   |              |   |              |+---------------+   +--------------+   +---------------+   +--------------+   +--------------+

class NumMatrix {public:    vector<vector<int>> sum;    int row,col;    NumMatrix(vector<vector<int>> matrix) {        row=matrix.size();        col=row>0?matrix[0].size():0;        sum=vector<vector<int>>(row+1,vector<int>(col+1,0));        for(vector<vector<int>>::size_type i=1;i<row+1;i++)        for(vector<int>::size_type j=1;j<col+1;j++)        {            sum[i][j]=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]+matrix[i-1][j-1];        }    }        int sumRegion(int row1, int col1, int row2, int col2) {    return sum[row2+1][col2+1]+sum[row1][col1]-sum[row1][col2+1]-sum[row2+1][col1];//注意新的部分和数组和原数组下标的关系，新数组下表要加一    }        };/** * Your NumMatrix object will be instantiated and called as such: * NumMatrix obj = new NumMatrix(matrix); * int param_1 = obj.sumRegion(row1,col1,row2,col2); */

注意：考虑传入数组为空的情况，定义了row和col，避免当数组为空时调用matrix[0].size()，导致数组越界