LeetCode

303. Range Sum Query - Immutable

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.

Example:

Given nums = [-2, 0, 3, -5, 2, -1]sumRange(0, 2) -> 1sumRange(2, 5) -> -1sumRange(0, 5) -> -3

Note:

1. You may assume that the array does not change.
2. There are many calls to sumRange function.

实现题，不知道题解写点啥好，创建的时间复杂度O(n)，求解的时间复杂度O(1)

创建数组时直接求了前缀和，这样求解的时候就可以O(1)求出答案了

class NumArray {public:    NumArray(vector<int> nums) {        num.push_back(0);        for (int i = 0; i < nums.size(); ++i) {            num.push_back(nums[i] + num[i]);        }    }        int sumRange(int i, int j) {        return num[j+1] - num[i];    }private:    vector<int> num;};/** * Your NumArray object will be instantiated and called as such: * NumArray obj = new NumArray(nums); * int param_1 = obj.sumRange(i,j); */

304. Range Sum Query 2D - Immutable

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) -> 8sumRegion(1, 1, 2, 2) -> 11sumRegion(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.

延续上一题思路，依然是求一个“前缀和”，不过这次求一个二维的。创建时间复杂度O(mn)，求解时间复杂度O(1)，空间复杂度都是O(mn)

num[i+1][j+1]表示matrix[0][0]~matrix[i][j]的和，则有下列关系

+-----+-+-------+     +--------+-----+     +-----+---------+     +-----+--------+|     | |       |     |        |     |     |     |         |     |     |        ||     | |       |     |        |     |     |     |         |     |     |        |+-----+-+       |     +--------+     |     |     |         |     +-----+        ||     | |       |  =  |              |  +  |     |         |  -  |              |+-----+-+       |     |              |     +-----+         |     |              ||               |     |              |     |               |     |              ||               |     |              |     |               |     |              |+---------------+     +--------------+     +---------------+     +--------------+    num[i][j]      =    num[i-1][j]    +     num[i][j-1]     -     num[i-1][j-1]   +     matrix[i-1][j-1]

因此，我们求返回值时，也有下列关系

+---------------+   +--------------+   +---------------+   +--------------+   +--------------+|               |   |         |    |   |   |           |   |         |    |   |   |          ||   (r1,c1)     |   |         |    |   |   |           |   |         |    |   |   |          ||   +------+    |   |         |    |   |   |           |   +---------+    |   +---+          ||   |      |    | = |         |    | - |   |           | - |      (r1,c2) | + |   (r1,c1)    ||   |      |    |   |         |    |   |   |           |   |              |   |              ||   +------+    |   +---------+    |   +---+           |   |              |   |              ||        (r2,c2)|   |       (r2,c2)|   |   (r2,c1)     |   |              |   |              |+---------------+   +--------------+   +---------------+   +--------------+   +--------------+

代码如下：

class NumMatrix {public:    NumMatrix(vector<vector<int>> matrix) {        int m = matrix.size();        int n = m > 0 ? matrix[0].size() : 0;        num = vector<vector<int> >(m+1, vector<int>(n+1, 0));        for (int i = 1; i <= m; ++i) {            for (int j = 1; j <= n; ++j) {                num[i][j] = num[i-1][j] + num[i][j-1] - num[i-1][j-1] + matrix[i-1][j-1];            }        }    }        int sumRegion(int row1, int col1, int row2, int col2) {        return num[row2+1][col2+1] - num[row1][col2+1] - num[row2+1][col1] + num[row1][col1];    }private:    vector<vector<int> > num;};/** * Your NumMatrix object will be instantiated and called as such: * NumMatrix obj = new NumMatrix(matrix); * int param_1 = obj.sumRegion(row1,col1,row2,col2); */