308. Range Sum Query 2D - Mutable --------H

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
update(3, 2, 2)
sumRegion(2, 1, 4, 3) -> 10

Note:

1. The matrix is only modifiable by the update function.
2. You may assume the number of calls to update and sumRegion function is distributed evenly.
3. You may assume that row1 ≤ row2 and col1 ≤ col2.

A:

就是建立一个2D数组。保存从原点到该位置的和

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 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44

class NumMatrix { public: NumMatrix(vector<vector<int>>& matrix) { m = matrix.size(); if(m==0){ return; } n = matrix[0].size(); M = vector<vector<long>>(m+1,vector<long>(n+1,0)); for(int i =1;i<=m;i++){ for(int j = 1; j<=n; j++){ M[i][j] = matrix[i-1][j-1] + M[i-1][j]+M[i][j-1] - M[i-1][j-1]; } } } void update(int row, int col, int val) { long orig = M[row+1][col+1] - M[row][col+1] - M[row+1][col] + M[row][col]; long diff = val - orig; for(int i = row+1; i<=m;i++){ for(int j = col+1; j<=n; j++){ M[i][j] += diff; } } } int sumRegion(int row1, int col1, int row2, int col2) { long downRight = M[row2+1][col2+1]; long topRight = M[row1][col2+1]; long downLeft = M[row2+1][col1]; long topLeft = M[row1][col1]; return int( downRight - topRight - downLeft + topLeft ); } private: vector<vector<long>> M; int m=0, n = 0; }; /** * Your NumMatrix object will be instantiated and called as such: * NumMatrix* obj = new NumMatrix(matrix); * obj->update(row,col,val); * int param_2 = obj->sumRegion(row1,col1,row2,col2); */

Note：
注意的一点是：求的row1 -1 , col-1的交叉
因为在constructor中没有检测matrix == null，耽误了1个小时，╮(╯▽╰)╭