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) -> 10Note:
-
The matrix is only modifiable by the update function.
-
You may assume the number of calls to update and sumRegion function is distributed evenly.
-
You may assume that row1 ≤ row2 and col1 ≤ col2.
分析：https://segmentfault.com/a/1190000004238792
注意这道题说明了"calls to update and sumRegion function is distributed evenly"。我们可以先不考虑这道题的限制，根据这个两个方法使用次数分情况讨论：
-
update多，sumRegion少
这种情况比较简单，我们可以直接复制矩阵，update的时候直接update对应点的值即可，sumRegion直接遍历指定范围内的值就可以了。update: O(1), sumRegion: O(mn) -
update少，sumRegion多。
我们可以不复制整个矩阵，而是在每个点处存(0, 0)到该的长方形内所有元素的和，这样的话，sumRegion的复杂度会很低，update的时候我们需要update所有受影响的sum。update: O(mn), sumRegion: O(1) -
update多，sumRegion多
（本题情况）
我们可以每个点存对于该行，起始点到该点为止的sum。这样话，update的话，我们只需要update该行受影响的sum即可。sumRegion的话，我们只需要遍历相应行，加上不同行的对应sum即可。update: O(n), sumRegion: O(m)
当然，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，尽量不搞复杂。
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)
1 public class NumMatrix { 2 int[][] rowSums; 3 4 public NumMatrix(int[][] matrix) { 5 if (matrix.length == 0) 6 return; 7 rowSums = new int[matrix.length][matrix[0].length]; 8 9 // 建rowSums矩阵 10 for (int i = 0; i < matrix.length; i++) { 11 for (int j = 0; j < matrix[0].length; j++) { 12 rowSums[i][j] = matrix[i][j] + (j == 0 ? 0 : rowSums[i][j - 1]); 13 } 14 } 15 } 16 17 public void update(int row, int col, int val) { 18 // 求出新值与旧值的差 19 int diff = val - (rowSums[row][col] - (col == 0 ? 0 : rowSums[row][col - 1])); 20 21 // 更新该行受影响的sum 22 for (int j = col; j < rowSums[0].length; j++) { 23 rowSums[row][j] += diff; 24 } 25 } 26 27 public int sumRegion(int row1, int col1, int row2, int col2) { 28 int res = 0; 29 30 // 逐行求和，每行的相应和为两sum相减 31 for (int i = row1; i <= row2; i++) { 32 res += rowSums[i][col2] - (col1 == 0 ? 0 :rowSums[i][col1 - 1]); 33 } 34 return res; 35 } 36 } 37 38 39 // Your NumMatrix object will be instantiated and called as such: 40 // NumMatrix numMatrix = new NumMatrix(matrix); 41 // numMatrix.sumRegion(0, 1, 2, 3); 42 // numMatrix.update(1, 1, 10); 43 // numMatrix.sumRegion(1, 2, 3, 4);
Range Sum Query 2D - Mutable & Immutable
就是没有update操作，这样的话，我们直接找出0,0 到每个点的和即可。
1 public class NumMatrix { 2 int[][] matrix; 3 4 public NumMatrix(int[][] m) { 5 matrix = m; 6 if (m == null || m.length == 0 || m[0].length == 0) return; 7 for (int i = 1; i < matrix[0].length; i++) { 8 matrix[0][i] += matrix[0][i - 1]; 9 } 10 11 for (int i = 1; i < matrix.length; i++) { 12 matrix[i][0] += matrix[i - 1][0]; 13 } 14 15 for (int i = 1; i < matrix.length; i++) { 16 for (int j = 1; j < matrix[0].length; j++) { 17 matrix[i][j] += matrix[i - 1][j] + matrix[i][j - 1] - matrix[i - 1][j - 1]; 18 } 19 } 20 } 21 22 public int sumRegion(int row1, int col1, int row2, int col2) { 23 if (matrix == null) return 0; 24 25 if (row1 == 0 && col1 == 0) return matrix[row2][col2]; 26 if (row1 == 0) return matrix[row2][col2] - matrix[row2][col1 - 1]; 27 if (col1 == 0) return matrix[row2][col2] - matrix[row1 - 1][col2]; 28 29 return matrix[row2][col2] - matrix[row1 - 1][col2] - matrix[row2][col1 - 1] + matrix[row1 - 1][col1 - 1]; 30 } 31 } 32 33 34 // Your NumMatrix object will be instantiated and called as such: 35 // NumMatrix numMatrix = new NumMatrix(matrix); 36 // numMatrix.sumRegion(0, 1, 2, 3); 37 // numMatrix.sumRegion(1, 2, 3, 4);
