[leetcode] 308. Range Sum Query 2D - Mutable 解题报告

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

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
update(3, 2, 2)
sumRegion(2, 1, 4, 3) -> 10

Note:

1. The matrix is only modifiable by the update function.
2. You may assume the number of calls to update and sumRegion function is distributed evenly.
3. You may assume that row1 ≤ row2 and col1 ≤ col2.

思路: 创建一个数组来标记每一行到前一个的和是多少, 这样求每一行从col1到col2的时候只要将其相减即可求出其和. 然后每次更新一个结点的时候, 要将我们的求和数组那一列之后的和求值.

代码如下:

class NumMatrix {
public:
    NumMatrix(vector<vector<int>> &matrix) {
        for(int i =0; i< matrix.size(); i++)
            hash.push_back(vector<int>(matrix[0].size()+1, 0));
        for(int i =0; i< matrix.size(); i++)
            for(int j =0; j< matrix[0].size(); j++)
                hash[i][j+1] = hash[i][j] + matrix[i][j];
    }

    void update(int row, int col, int val) {
        int diff =val - (hash[row][col+1]-hash[row][col]);
        for(int i = col+1; i < hash[row].size(); i++)
            hash[row][i] += diff;
    }

    int sumRegion(int row1, int col1, int row2, int col2) {
        int sum = 0;
        for(int i = row1; i<= row2; i++)
            sum += (hash[i][col2+1] - hash[i][col1]);
        return sum;
    }
private:
    vector<vector<int>> hash;
};


// Your NumMatrix object will be instantiated and called as such:
// NumMatrix numMatrix(matrix);
// numMatrix.sumRegion(0, 1, 2, 3);
// numMatrix.update(1, 1, 10);
// numMatrix.sumRegion(1, 2, 3, 4);