【刷题-LeetCode】304. Range Sum Query 2D

304. 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.

解 同一维数组一样，先预处理

定义数组(mathrm{S}[i][j])表示前 i-1 行前 j-1 列交叉区域的和

• 预处理阶段：(mathrm{S}[i][j] =mathrm{M}[i-1][j-1]mathrm{S}[i][j-1]+mathrm{S}[i-1][j] - mathrm{S}[i-1][j-1])

• 查询阶段：(mathrm{sum\_of\_region}[r1, c1, r2, c2] = mathrm{S}[r2+1][c2+1]-mathrm{S}[r1][c2+1]-mathrm{S}[r2+1][c1]+mathrm{S}[r1][c1])

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