304. Range Sum Query 2D - Immutable

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本文介绍了一种高效解决二维矩阵中指定矩形区域元素之和的方法。提供了两种解决方案：一种是对每行预处理求和，查询时计算所需区间；另一种是使用累加矩阵技巧，实现常数时间查询。

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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.

题解

求两个点之间的矩形中所有数的和

解法1

构造：对每个点找出所在行从0到该点的和，用时O(m*n)
查询：从row1到row2，通过相减找出col1到col2的和，用时O(m)

class NumMatrix {
public:
    NumMatrix(vector<vector<int>> matrix) {
        int sum;
        haha = matrix;
        for(int i = 0; i < matrix.size(); i++){
            for(int j = 1; j < matrix[0].size(); j++){
                haha[i][j] += haha[i][j-1];
            }
        }
    }

    int sumRegion(int row1, int col1, int row2, int col2) {
        int sum = 0;
        for(int i = row1; i <= row2; i++){
            if(col1)    sum += haha[i][col2] - haha[i][col1-1];
            else    sum += haha[i][col2];
        }
        return sum;
    }
private:
    vector<vector<int>> haha;
};

解法2

构造：

能不能找到从(0,0)到任意点(i,j)的矩形中所有数的和呢？
可以，有状态转移方程
sums[i][j] = sums[i-1][j] + sums[i][j-1] - sums[i-1][j-1] + matrix[i][j]。

为了避免边界判断增加sums[0][0：col]={0}和sums[0：row][0]={0}，所以sums[i+1][j+1]表示从（0，0）到（i，j）的矩形中数的和。
此时状态转移方程为
sums[i][j] = sums[i-1][j] + sums[i][j-1] - sums[i-1][j-1] + matrix[i-1][j-1]。

用时O(m*n)

查询：

此时（row1，col1），（row2，col2）围成的矩形中数的和就等于
sums[row2+1][col2+1] - sums[row2+1][col1] - sums[row1][col2+1] + sums[row1][col1]
用时O(1)

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

    int sumRegion(int row1, int col1, int row2, int col2) {
        return haha[row2+1][col2+1] - haha[row2+1][col1] - haha[row1][col2+1] + haha[row1][col1];
    }
private:
    vector<vector<int>> haha;
};