Range Sum Query - Immutable

题目：

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:

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

代码：

class NumArray {
public:

int *numbers;
NumArray(vector<int> &nums) {
if(nums.size()!=0)
{
numbers=new int[nums.size()];
numbers[0]=nums[0];
for(int i=1; i<nums.size(); i++)
{
numbers[i]=numbers[i-1]+nums[i];
}
}

}

int sumRange(int i, int j) {
int sum=0;
if(i==0)
{
sum=numbers[j]-numbers[i];
sum+=(numbers[0]);
}
else
{
sum=numbers[j]-numbers[i-1];
}
return sum;

}
};

// Your NumArray object will be instantiated and called as such:
// NumArray numArray(nums);
// numArray.sumRange(0, 1);
// numArray.sumRange(1, 2);