[LeetCode]Range Sum Query 2D - Mutable

Range Sum Query 2D - Mutable

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.

分析

注意这道题说明了"calls to update and sumRegion function is distributed evenly"。咱们能够先不考虑这道题的限制，根据这个两个方法使用次数分状况讨论：

• update多，sumRegion少
  这种状况比较简单，咱们能够直接复制矩阵，update的时候直接update对应点的值便可，sumRegion直接遍历指定范围内的值就能够了。

  update: O(1), sumRegion: O(mn)

• update少，sumRegion多。
  咱们能够不复制整个矩阵，而是在每一个点处存(0, 0)到该的长方形内全部元素的和，这样的话，sumRegion的复杂度会很低，update的时候咱们须要update全部受影响的sum。

  update: O(mn), sumRegion: O(1)

• update多，sumRegion多（本题状况）
  咱们能够每一个点存对于该行，起始点到该点为止的sum。这样话，update的话，咱们只须要update该行受影响的sum便可。sumRegion的话，咱们只须要遍历相应行，加上不一样行的对应sum便可。

  update: O(n), sumRegion: O(m)

固然，对于这种类型的题目，使用一些高级数据结构会更时间复杂度会更低，能达到logn，如二维线段树。这里只涉及基本的数据结构，尽可能不搞复杂。

复杂度:

注：m指行数，n指列数，这里global的矩阵不算各个方法的extra space。

update

time: O(n), space: O(1)

sumRegion

time: O(m), space: O(1)

代码

public class NumMatrix {
    int[][] rowSums;
    
    public NumMatrix(int[][] matrix) {
        if (matrix.length == 0)
            return;
        rowSums = new int[matrix.length][matrix[0].length];
        
        // 建rowSums矩阵
        for (int i = 0; i < matrix.length; i++) {
            for (int j = 0; j < matrix[0].length; j++) {
                rowSums[i][j] = matrix[i][j] + (j == 0 ? 0 : rowSums[i][j - 1]);
            }
        }
    }

    public void update(int row, int col, int val) {
        // 求出新值与旧值的差
        int diff = val - (rowSums[row][col] - (col == 0 ? 0 : rowSums[row][col - 1]));
        
        // 更新该行受影响的sum
        for (int j = col; j < rowSums[0].length; j++) {
            rowSums[row][j] += diff;
        }
    }

    public int sumRegion(int row1, int col1, int row2, int col2) {
        int res = 0;

        // 逐行求和，每行的相应和为两sum相减
        for (int i = row1; i <= row2; i++) {
            res += rowSums[i][col2] - (col1 == 0 ? 0 :rowSums[i][col1 - 1]);
        }
        return res;
    }
}


// Your NumMatrix object will be instantiated and called as such:
// NumMatrix numMatrix = new NumMatrix(matrix);
// numMatrix.sumRegion(0, 1, 2, 3);
// numMatrix.update(1, 1, 10);
// numMatrix.sumRegion(1, 2, 3, 4);