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304.range-sum-query-2d-immutable.cpp
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72 lines (70 loc) · 1.98 KB
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/*
* @lc app=leetcode id=304 lang=cpp
*
* [304] Range Sum Query 2D - Immutable
*
* https://leetcode.com/problems/range-sum-query-2d-immutable/description/
*
* algorithms
* Medium (32.85%)
* Total Accepted: 75.1K
* Total Submissions: 226.1K
* Testcase Example: '["NumMatrix","sumRegion","sumRegion","sumRegion"]\n[[[[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]]],[2,1,4,3],[1,1,2,2],[1,2,2,4]]'
*
* 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.
*
*
*/
class NumMatrix {
public:
NumMatrix(vector<vector<int>>& matrix) {
if (matrix.empty() || matrix[0].empty())
return;
const int m = matrix.size(), n = matrix[0].size();
sum = vector<vector<int>>(m + 1, vector<int>(n + 1, 0));
for (int i = 1; i <= m; ++i)
for (int j = 1; j <= n; ++j)
sum[i][j] = sum[i-1][j] + sum[i][j-1] - sum[i-1][j-1] + matrix[i-1][j-1];
return;
}
int sumRegion(int row1, int col1, int row2, int col2) {
return sum[row2+1][col2+1] + sum[row1][col1] - sum[row2+1][col1] - sum[row1][col2+1];
}
private:
vector<vector<int>> sum;
};
/**
* Your NumMatrix object will be instantiated and called as such:
* NumMatrix* obj = new NumMatrix(matrix);
* int param_1 = obj->sumRegion(row1,col1,row2,col2);
*/