[leetcode] 304. Range Sum Query 2D - Immutable 解题报告

题目链接：https://leetcode.com/problems/range-sum-query-2d-immutable/

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:

1. You may assume that the matrix does not change.
2. There are many calls to sumRegion function.
3. You may assume that row1 ≤ row2 and col1 ≤ col2.

思路：按照直接的求前n项和的思路来做这道题就可以过, 当然也可以用二叉索引树来做.

代码如下:

class NumMatrix {
public:
    NumMatrix(vector> &matrix) {
        if(matrix.size()==0) return;
        int row = matrix.size(), col = matrix[0].size();
        sum.resize(row, vector(col+1, 0));
        for(int i =0; i < row; i++)
            for(int j =0; j < col; j++) 
                sum[i][j+1] = sum[i][j] + matrix[i][j];
    }
    
    int sumRegion(int row1, int col1, int row2, int col2) {
        int ans = 0;
        for(int i = row1; i <= row2; i++)
            ans += sum[i][col2+1] - sum[i][col1];
        return ans;
    }
private:
    vector> sum;
};


// Your NumMatrix object will be instantiated and called as such:
// NumMatrix numMatrix(matrix);
// numMatrix.sumRegion(0, 1, 2, 3);
// numMatrix.sumRegion(1, 2, 3, 4);