304. Range Sum Query 2D - Immutable

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本文介绍了一种高效计算二维矩阵中任意指定矩形区域元素之和的方法。通过预先计算并存储累积和，使得每次查询时间复杂度降低，适用于需要频繁进行区间和查询的应用场景。

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题目：

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).

Range Sum Query 2D
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:

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

题意：

给定一个整型二维数组matrix，找到由给定的左上角元素与右下角元素所构成的矩形部分所有整数的和。

note：

1、可以假定该矩阵不会改变；

2、sumRegion函数会被多次调用；

3、可以假定row1<=row2，col1<=col2；

思路：

与Range Sum Query - Immutable 区域和检索类似，维护一个二维数组sums，其中sums[i][j]表示累计区间（0,0）到（i,j）这个矩形区间所有的数字之和，此时需要快速求出（r1,c1）到（r2,c2）的矩形区域的累计值时，只需sums[r2][c2] - sums[r2][c1-1] - sums[r1-1][c2] + sums[r1-1][c1-1]即可。

代码：C++版：28ms

class NumMatrix {
private:
    int row, col;
    vector<vector<int>> sums;
public:
    NumMatrix(vector<vector<int>> &matrix) {
        row = matrix.size();
        col = row>0 ? matrix[0].size() : 0;
        sums = vector<vector<int>>(row+1, vector<int>(col+1, 0));
        for(int i=1; i<=row; i++) {
            for(int j=1; j<=col; j++) {
                sums[i][j] = matrix[i-1][j-1] + sums[i-1][j] + sums[i][j-1] - sums[i-1][j-1] ;
            }
        }
    }


    int sumRegion(int row1, int col1, int row2, int col2) {
        return sums[row2+1][col2+1] - sums[row2+1][col1] - sums[row1][col2+1] + sums[row1][col1];
    }
};




// 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);