Leetcode第303 Range Sum Query - Immutable（简单动态规划解法）

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.
实现代码如下：

class NumArray{
public:
NumArray(vector<int> &nums){
if (!nums.size())
return;
sum.push_back(nums[0]);
for (int i = 1; i < nums.size(); i++){
sum.push_back(nums[i] + sum[i - 1]);
}
}

int sumRange(int i, int j){
if (i == 0)
return sum[j];
else
return (sum[j] - sum[i-1]);
}
private:
vector<int> sum;
};