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 同一維陣列一樣，先預處理

定義陣列\(\mathrm{S}[i][j]\)表示前 i-1 行前 j-1 列交叉區域的和

• 預處理階段：\(\mathrm{S}[i][j] =\mathrm{M}[i-1][j-1]\mathrm{S}[i][j-1]+\mathrm{S}[i-1][j] - \mathrm{S}[i-1][j-1]\)

• 查詢階段：\(\mathrm{sum\_of\_region}[r1, c1, r2, c2] = \mathrm{S}[r2+1][c2+1]-\mathrm{S}[r1][c2+1]-\mathrm{S}[r2+1][c1]+\mathrm{S}[r1][c1]\)

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];
    }
};