304. Range Sum Query 2D

问题：

给定一个二维数组，

求给定区间围成矩形的和。

Example 1:
Input
["NumMatrix", "sumRegion", "sumRegion", "sumRegion"]
[[[[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]]], [2, 1, 4, 3], [1, 1, 2, 2], [1, 2, 2, 4]]
Output
[null, 8, 11, 12]

Explanation
NumMatrix numMatrix = new NumMatrix([[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]]);
numMatrix.sumRegion(2, 1, 4, 3); // return 8 (i.e sum of the red rectangele).
numMatrix.sumRegion(1, 1, 2, 2); // return 11 (i.e sum of the green rectangele).
numMatrix.sumRegion(1, 2, 2, 4); // return 12 (i.e sum of the blue rectangele).
 

Constraints:

m == matrix.length
n == matrix[i].length
1 <= m, n <= 200
-10^5 <= matrix[i][j] <= 10^5
0 <= row1 <= row2 < m
0 <= col1 <= col2 < n
At most 10^4 calls will be made to sumRegion.

　　

example 1:

解法：DP，preSum

1. 状态：dp[i][j]：[0,0]~[i-1,j-1]的矩形和

• i：i坐标
• j：j坐标

2.选择：

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

3.base:

• dp[0][j]=0
• dp[i][0]=0

而题目所求矩形[i1,j1]~[i2,j2]的和为：

• res=dp[i2][j2]-dp[i2][j1]-dp[i1][j2]+dp[i1][j1]

代码参考：

class NumMatrix {
public:
    vector<vector<int>> dp;
    NumMatrix(vector<vector<int>>& matrix) {
        int n = matrix.size();
        int m = matrix[0].size();
        dp.resize(n+1, vector<int>(m+1, 0));
        for(int i=1; i<=n; i++) {
            for(int j=1; j<=m; j++) {
                dp[i][j]=dp[i][j-1]+dp[i-1][j]-dp[i-1][j-1]+matrix[i-1][j-1];
            }
        }
        return;
    }
    
    int sumRegion(int row1, int col1, int row2, int col2) {
        return dp[row2+1][col2+1]-dp[row2+1][col1]-dp[row1][col2+1]+dp[row1][col1];
    }
};

/**
 * Your NumMatrix object will be instantiated and called as such:
 * NumMatrix* obj = new NumMatrix(matrix);
 * int param_1 = obj->sumRegion(row1,col1,row2,col2);
 */