LeetCode|Range Sum Query 2D - Immutable

Range Sum Query 2D - Immutable

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

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：

在Range Sum Query题目中我们预先计算出行中[0， i]， i∈（0， size-1）的和。此题思路相同，我预先计算出每一行[0， i]， i∈（0， row）的和，然后再计算出[(0, 0), (i, j)], i∈（0， row）, j∈（0， col）的面积，用的都是动态规划的思想。

为什么不预先计算每一列的和呢？其实也是可以的，但是根据我目前的理解，vector < vector >还是优先遍历行再遍历列的效率会更高。

class NumMatrix {
public:
NumMatrix(vector<vector<int> > &matrix) {
if(matrix.size() == 0) return;
int row = matrix.size(), col = matrix[0].size();
regionSum.resize(row);

for(int i = 0; i < row; i++){  // calc the sum of row
regionSum[i].resize(col);
regionSum[i][0] = matrix[i][0];
for(int j = 1; j < col; j++){
regionSum[i][j] = regionSum[i][j-1] + matrix[i][j];
}
}
for(int i = 1; i < row; i++){  // calc the sum of region
for(int j = 0; j < col; j++){
regionSum[i][j] += regionSum[i-1][j];
}
}
}

int sumRegion(int row1, int col1, int row2, int col2) {
if(row1 > row2 || col1 > col2) return 0;
int res = regionSum[row2][col2];
int state = 0;
if(row1){
res -= regionSum[row1-1][col2];
state++;
}
if(col1){
res -= regionSum[row2][col1-1];
state++;
}
if(state == 2) res += regionSum[row1-1][col1-1];
return res;
}
private:
vector<vector<int> > regionSum;
};

// 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);