Leetcode解题: Range Sum Query 2D - Immutable

发表于 | 分类于

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).

alt text
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.

解法1：DP: O(N^2) Initialization, O(1) Function Call

这题不难，题目意思是里面的程序会要call很多次。如果不做优化的话每次的成本都是O(N^2).
如果我们先预处理一下，把每一个点的累计和存储一下，就是说dp[i][j]是从[0][0]到[i][j]的和。
那么当要计算子矩阵的和的时候，我们可以得到如下的关系

1
sum[i][j] = dp[i][j] - dp[i][j-1] - dp[i-1][j] + dp[i-1][j-1]

C++

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
class NumMatrix {
public:
    NumMatrix(vector<vector<int>> matrix) {
        // initialize dp
        dp = matrix;    // copy assignment operator
        for (int i = 0; i < matrix.size(); ++i) {
            for (int j = 0; j < matrix[i].size(); ++j) {
                dp[i][j] = (i == 0?0: dp[i - 1][j]) + (j == 0?0:dp[i][j - 1]) 
                - (i > 0 && j > 0?dp[i -1][j - 1]: 0) + matrix[i][j];
            }
        }
    }
    
    int sumRegion(int row1, int col1, int row2, int col2) {
        return dp[row2][col2] - (row1 > 0? dp[row1 - 1][col2]: 0) - (col1 > 0 ? dp[row2][col1 - 1]: 0) + (row1 > 0 && col1 >0 ? dp[row1 - 1][col1 - 1]: 0);
    }
    
private:
    vector<vector<int>> dp;
};
/**
 * Your NumMatrix object will be instantiated and called as such:
 * NumMatrix obj = new NumMatrix(matrix);
 * int param_1 = obj.sumRegion(row1,col1,row2,col2);
 */

Java

1