[Leetcode] 304. Range Sum Query 2D - Immutable 解题报告

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).



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.
思路：

和上一道题目的思路一样，只不过这次是变成了二维的情况：我们用sums[i+1][j+1]表示从(0,0)到(i, j)这个矩形之内的数组元素之和：

1）计算sums数组：sums[i+1][j+1] = sums[i][j+1] + sums[i+1][j] - sums[i][j] + matrix[i][j]；时间复杂度是O(mn)，其中m和n分别是matrix的行和列个数。

2）计算矩形和：由(row1, col1)到(row2, col2)构成的矩形的和的计算公式是：sumRegion(row1, col1, row2, col2) = sums[row2+1][col2+1] - sums[row1][col2+1] - sums[row2+1][col1] + sums[row1][col1]。时间复杂度是O(1)。

该算法的空间复杂度是O(mn)。时间复杂度应该是对于本题目而言最优的了。

代码：

class NumMatrix {
public:
NumMatrix(vector<vector<int>> matrix) {
if (matrix.size() == 0 || matrix[0].size() == 0) {
return;
}
int row_num = matrix.size(), col_num = matrix[0].size();
sums = vector<vector<int>>(row_num + 1, vector<int>(col_num + 1, 0));
for(int i = 0; i < row_num; ++i) {
for (int j = 0; j < col_num; ++j) {
sums[i + 1][j + 1] = sums[i][j + 1] + sums[i + 1][j] - sums[i][j] + matrix[i][j];
}
}
}

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

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