leetcode 304. Range Sum Query 2D

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04. Range Sum Query 2D - Immutable

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Total Accepted: 12544 Total Submissions: 56503 Difficulty: Medium

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:

1. You may assume that the matrix does not change.
2. There are many calls to sumRegion function.
3. You may assume that row1 ≤ row2 and col1 ≤ col2.

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Submission Details

12 / 12 test cases passed.	Status:  Accepted
Runtime: 24 ms

思路：动态规划

 

class NumMatrix {
public:
    NumMatrix(vector<vector<int>> &matrix) {
        if(matrix.empty()){
            return;
        }
        int n = matrix.size(), m = matrix[0].size();
        sums = vector<vector<int>>(n+1, vector<int>(m+1, 0));
        int i,j;
        sums[0][0] = matrix[0][0];
        for(i = 1;i <= n;i++){
            for(j = 1;j <= m;j++){
                sums[i][j] = sums[i - 1][j] + sums[i][j - 1] - sums[i - 1][j - 1] + matrix[i -1][j - 1];
            }
        }
    }

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

private:    
    vector<vector<int>> sums;
};


// 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);