剑指 Offer II 013. 二维子矩阵的和

给定一个二维矩阵 matrix，以下类型的多个请求：

    计算其子矩形范围内元素的总和，该子矩阵的左上角为 (row1, col1) ，右下角为 (row2, col2) 。

实现 NumMatrix 类：

    NumMatrix(int[][] matrix) 给定整数矩阵 matrix 进行初始化
    int sumRegion(int row1, int col1, int row2, int col2) 返回左上角 (row1, col1) 、右下角 (row2, col2) 的子矩阵的元素总和。

 

示例 1：

 

 

输入:
["NumMatrix","sumRegion","sumRegion","sumRegion"]
[[[[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]]],[2,1,4,3],[1,1,2,2],[1,2,2,4]]
输出:
[null, 8, 11, 12]

解释:
NumMatrix numMatrix = new NumMatrix([[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]]]);
numMatrix.sumRegion(2, 1, 4, 3); // return 8 (红色矩形框的元素总和)
numMatrix.sumRegion(1, 1, 2, 2); // return 11 (绿色矩形框的元素总和)
numMatrix.sumRegion(1, 2, 2, 4); // return 12 (蓝色矩形框的元素总和)

 

提示：

    m == matrix.length
    n == matrix[i].length
    1 <= m, n <= 200
    -105 <= matrix[i][j] <= 105
    0 <= row1 <= row2 < m
    0 <= col1 <= col2 < n
    最多调用 104 次 sumRegion 方法

 解析：

　　对二维矩阵求一个前缀和，每次query减去上边的和左边的即可，注意有重叠

class NumMatrix {
public:
    vector<vector<int>> matrix;
    int presum[210][210];
    int n, m;
    NumMatrix(vector<vector<int>>& matrix) {
        this->matrix = matrix;
        int n = matrix.size();
        int m = matrix[0].size();
        this->n = n;
        this->m = m;
        for(int i = 0; i < n; i++)
        {
            int rsum = 0;
            for(int j = 0; j < m; j++)
            {
                if(i == 0 && j == 0)
                    presum[i][j] = matrix[i][j];
                else if(i == 0)
                    presum[i][j] = presum[i][j - 1] + matrix[i][j];
                else
                {
                    rsum += matrix[i][j];
                    presum[i][j] = presum[i - 1][j] + rsum;
                }
            }
        }

    }
    
    int sumRegion(int row1, int col1, int row2, int col2) {
        int temp =presum[row2][col2];
        int temp1 = 0, temp2 = 0;
        if(row1 - 1 >= 0)
        {
            temp1 = presum[row1 - 1][col2];

        }
        if(col1 - 1 >= 0)
        {
            if(row1 - 1 >= 0)
                temp2 = presum[row2][col1 - 1] - presum[row1 - 1][col1 - 1];
            else
                temp2 = presum[row2][col1 - 1];
        }
        return temp - temp1 - temp2;

    }
};

/**
 * Your NumMatrix object will be instantiated and called as such:
 * NumMatrix* obj = new NumMatrix(matrix);
 * int param_1 = obj->sumRegion(row1,col1,row2,col2);
 */