lintcode Range Sum Query 2D - Immutable

Range Sum Query 2D - Immutable 

 描述
 笔记

 数据

 评测

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)

.


 注意事项

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.

您在真实的面试中是否遇到过这个题？ 

哪家公司问你的这个题？
感谢您的反馈

样例

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

我自己写的算法是逐个相加，测试通过但是时间花费为800ms，以下为我的代码

class NumMatrix {
public:
NumMatrix(vector<vector<int>> ma) :matrix(ma){

}

int sumRegion(int row1, int col1, int row2, int col2) {
int r1=row1,c1,r2=row2,c2=col2;
int sum=0;
for(;r1<=r2;r1++)
{
c1=col1;
for(;c1<=c2;c1++)
{
sum+=matrix[r1][c1];
}
}
return sum;
}
private:vector<vector<int>>matrix;
};

/* Your NumMatrix object will be instantiated and called as such:

 * NumMatrix obj = new NumMatrix(matrix);
 * int param_1 = obj.sumRegion(row1,col1,row2,col2);
 */

我看的别人的代码，测试时间为300ms，比我的快一倍，想不明白都是O（n2）的时间复杂度，只不过它的计算是在构造函数中，先记录下来

class NumMatrix {
private:
vector<vector<int>> dp;

public:
NumMatrix(vector<vector<int>> matrix) {
if
4000
(matrix.size() == 0 || matrix[0].size() == 0) {
return;
}
int n = matrix.size();
int m = matrix[0].size();

dp.resize(n + 1, vector<int>(m + 1, 0));
for (int r = 0; r < n; r++) {
for (int c = 0; c < m; c++) {
dp[r + 1][c + 1] = dp[r + 1][c] + dp[r][c + 1] + matrix[r][c] - dp[r][c];
}
}
}

int sumRegion(int row1, int col1, int row2, int col2) {
return dp[row2 + 1][col2 + 1] - dp[row1][col2 + 1] - dp[row2 + 1][col1] + dp[row1][col1];
}
};