303. Range Sum Query - Immutable

#week6

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) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3

 

Note:

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

 

分析：

动态规划类型

状态转移方程：f[i] = f[i-1] + nums[i];

 

代码：

超时版本：

 1 class NumArray {
 2 public:
 3     int** f;
 4     NumArray(vector<int> nums) {
 5         int len = nums.size();
 6         if (len > 0) {
 7         f = new int* [len];
 8         for (int i = 0; i < len; i++) f[i] = new int[len];
 9         f[0][0] = nums[0];
10         for (int i = 0; i < len; i++)
11             for (int j = i; j < len; j++) {
12                 if (i == j) f[i][j] = nums[i];
13                 else {
14                     f[i][j] = f[i][j-1] + nums[j];
15                 }
16             }
17         } else {
18             f = new int*[1];
19             f[0] = new int[1];
20             f[0][0] = 0;
21         }
22     }
23     
24     int sumRange(int i, int j) {
25         return f[i][j];
26     }
27 };
28 
29 /**
30  * Your NumArray object will be instantiated and called as such:
31  * NumArray obj = new NumArray(nums);
32  * int param_1 = obj.sumRange(i,j);
33  */

 

修改为一维，不超时：

 1 class NumArray {
 2 public:
 3     int* f;
 4     NumArray(vector<int> nums) {
 5         int len = nums.size();
 6         if (len > 0) {
 7             f = new int [len];
 8             f[0] = nums[0];
 9             for (int i = 1; i < len; i++)
10                 f[i] = f[i-1] + nums[i];
11         } else {
12             f = new int[1];
13             f[0] = 0;
14         }
15     }
16     
17     int sumRange(int i, int j) {
18         return f[j] - f[i-1];
19     }
20 };
21 
22 /**
23  * Your NumArray object will be instantiated and called as such:
24  * NumArray obj = new NumArray(nums);
25  * int param_1 = obj.sumRange(i,j);
26  */

 

posted @ 2018-01-13 02:23  小预备  阅读(143)  评论(0)  编辑  收藏  举报