Range Sum Query

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.

 

TIME LIMITED EXCEEDED VERSION

 1 class NumArray {
 2     private:
 3     vector<int> arr;
 4 public:
 5     NumArray(vector<int> &nums) {
 6         arr.clear();
 7         for(int i = 0; i < nums.size(); i++)
 8             arr.push_back(nums[i]);
 9     }
10 
11     int sumRange(int i, int j) {
12         i = i < 0 ? 0 : i;
13         j = j >= arr.size() ? arr.size() - 1 : j;
14         
15         int sum = 0;
16         for(int k = i; k <= j; k++)
17             sum += arr[k];
18         return sum;
19     }
20 };
21 
22 
23 // Your NumArray object will be instantiated and called as such:
24 // NumArray numArray(nums);
25 // numArray.sumRange(0, 1);
26 // numArray.sumRange(1, 2);

 

 

Runtime: 588ms

 1 class NumArray {
 2 public:
 3     NumArray(vector<int> &nums) {
 4         arr.push_back(0);
 5         for(int i = 0; i < nums.size(); i++)
 6             arr.push_back(arr[i] + nums[i]);
 7     }
 8 
 9     int sumRange(int i, int j) {
10         return arr[j + 1] - arr[i];
11     }
12 private: 
13     vector<int> arr;
14 };
15 
16 
17 // Your NumArray object will be instantiated and called as such:
18 // NumArray numArray(nums);
19 // numArray.sumRange(0, 1);
20 // numArray.sumRange(1, 2);

 

posted @ 2015-11-10 23:31  amazingzoe  阅读(136)  评论(0)  编辑  收藏  举报