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) -> 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.

题目解答：这个题目说了，会反复的求一段数字的和，但是默认数组不变。这么说来，就是求和的复杂度就是主要的复杂度。那么可以先把它们存起来，然后再求得时候，直接把结果返回来。需要注意的是，求[2,5]的和就是求[0，5]的和减去[0,2]的和，再加上[2]这个位置的值。

代码如下：

class NumArray {

public:

    NumArray(vector<int> &nums):nums(nums) {

        int sum = 0;

        for(int i = 0;i < nums.size();i++)

        {

            sum += nums[i];

            sums.push_back(sum);

        }

    }

    int sumRange(int i, int j) {

        if((i > j) || (i >= nums.size()) || (j >= nums.size()))

            return 0;

        return sums[j] - sums[i] + nums[i];

    }

private:

    vector<int> sums;

    vector<int> &nums;

};

// Your NumArray object will be instantiated and called as such:

// NumArray numArray(nums);

// numArray.sumRange(0, 1);

// numArray.sumRange(1, 2);