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 ( row 1, col 1) and lower right corner ( row 2, col 2).

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:

  1. You may assume that the matrix does not change.
  2. There are many calls to sumRegion function.
  3. You may assume that row 1 ≤ row 2 and col 1 ≤ col 2.

 

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,下面的代码中我们由于用了辅助列和辅助行,所以下标会有些变化,参见代码如下:

 

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;
};

 

类似题目:

Range Sum Query 2D - Mutable

Range Sum Query - Immutable

Range Sum Query - Mutable

 

LeetCode All in One 题目讲解汇总(持续更新中…)

 


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