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).

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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.

解题之法

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

分析

这道题让我们求一个二维区域和的检索，是之前那道题Range Sum Query - Immutable 区域和检索的延伸。
有了之前那道题的基础，我们知道这道题其实也是换汤不换药，还是要建立一个累计区域和的数组，然后根据边界值的加减法来快速求出给定区域之和。这里我们维护一个二维数组dp，其中dp[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，那么此时如果我们想要快速求出(r1, c1)到(r2, c2)的矩形区间时，只需dp[r2][c2] - dp[r2][c1 - 1] - dp[r1 - 1][c2] + dp[r1 - 1][c1 - 1]即可。