LeetCode #304 - Range Sum Query 2D - Immutable-优快云博客

本文链接：https://blog.youkuaiyun.com/LawFile/article/details/81195293

本文介绍了一种高效处理二维矩阵中任意矩形区域元素之和的方法。通过预处理矩阵生成一个辅助矩阵(dp)，使得后续对任意指定矩形区域的求和操作能在常数时间内完成。该方法适用于矩阵数据不变且需要频繁进行区间和查询的应用场景。

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题目描述：

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.

给定一个矩阵，求某一矩形区域的元素之和。

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