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

本文链接：https://blog.youkuaiyun.com/xinqrs01/article/details/55187277

本文介绍了一种高效解决二维矩阵区间求和问题的方法。通过预处理实现快速查询任意指定矩形区域内的元素总和，特别适用于矩阵不变且需频繁查询的情况。

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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).
这里写图片描述
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.

s思路:
1. 从1d变成2d，方法并没变，仍然通过o(m*n)预处理得到所有从[0,0]到[i,j]的所有元素之和！
2. bug:加括号，尤其是问号表达式！不然运算顺序就不保证了。


class NumMatrix {
private:
    vector<vector<int>> res;
public:
    NumMatrix(vector<vector<int>> matrix) {
        res=matrix;
        for(int i=0;i<res.size();i++){
            for(int j=0;j<res[0].size();j++){
                res[i][j]=res[i][j]+(i>0?res[i-1][j]:0)+(j>0?res[i][j-1]:0)-(i>0&&j>0?res[i-1][j-1]:0);//bug:加括号，尤其是问号表达式！不然运算顺序就不保证了。
            }    
        }
    }

    int sumRegion(int row1, int col1, int row2, int col2) {
        return res[row2][col2]-(col1>0?res[row2][col1-1]:0)-(row1>0?res[row1-1][col2]:0)+(col1>0&&row1>0?res[row1-1][col1-1]:0);//bug:加括号，尤其是问号表达式！不然运算顺序就不保证了。
    }
};

/**
 * Your NumMatrix object will be instantiated and called as such:
 * NumMatrix obj = new NumMatrix(matrix);
 * int param_1 = obj.sumRegion(row1,col1,row2,col2);
 */