[leetcode] 304. Range Sum Query 2D - Immutable
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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).
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
题意:
给一个二维的数组,找出某个局部的元素之和。元素的最上角是 (row1, col1),右下角是(row2, col2)。
思路:
我们可以这样子来计算指定区域的面积,这个指定的区域可以使用几个区域的运算来得到。这里的区域都是以(row,col)为右下角,(0,0)为左上角。那么(row1, col1),(row2,col2)这个区域其实就是(row2,col2)减去(row2,col1),(row1,col2)再加上(row1,col1)得到。
而开始计算以(row,col)为右下角的指定区域的元素和时,也可以时使用之前计算得到的结果来完成计算,(row,col)=(ro1-1,cow)+(row,col-1)+num[row][col]-(row-1,col-1)。
以上。
代码如下:
class NumMatrix {public: NumMatrix(vector<vector<int>> &matrix) { if(matrix.empty())return; sum.resize(matrix.size() + 1, vector<int>(matrix[0].size() + 1, 0)); for(auto i = 1; i < matrix.size() + 1; i++) for(auto j = 1; j < matrix[0].size() + 1; j++) sum[i][j] = sum[i-1][j] + sum[i][j-1] + matrix[i-1][j-1] - sum[i-1][j-1]; } int sumRegion(int row1, int col1, int row2, int col2) { return sum[row2 + 1][col2 + 1] - sum[row1][col2 + 1] - sum[row2 + 1][col1] + sum[row1][col1]; }private: vector<vector<int>> sum;};// Your NumMatrix object will be instantiated and called as such:// NumMatrix numMatrix(matrix);// numMatrix.sumRegion(0, 1, 2, 3);// numMatrix.sumRegion(1, 2, 3, 4);
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