leetcode 304. Range Sum Query 2D - Immutable（递推）

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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.

题解：

二维的情况和一维的思路类似，就是求一个递推式然后把和预处理出来，然后O(1)查询。

一维的递推式是：s[i]=s[i-1]+a[i];

二维的由画图可知：s[i][j]=a[i][j]+s[i-1][j]+s[i][j-1]-s[i-1][j-1];

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

    int sumRegion(int row1, int col1, int row2, int col2) {
        return s[row2+1][col2+1]-s[row2+1][col1]-s[row1][col2+1]+s[row1][col1];
    }
private:
    int n,m;
    vector<vector<int> >s;
};


// 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);

leetcode 304. Range Sum Query 2D - Immutable（递推）

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原文地址：http://www.cnblogs.com/zywscq/p/5410723.html