Range Sum Query

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

解答

这道题目是一道很简单的动态规划问题中的子结构重复问题，主要在于sumRange（）中如何计算；这里我采用先将所有（0，i）；i <nums.size()的子集储存在acc数组中，然后所有的计算都可以直接调用acc 数组相减完成计算；

具体代码如下：

class NumArray {public:    NumArray(vector<int> nums) {        accu.push_back(0);        for (iter = nums.begin(); iter != nums.end(); iter++)            accu.push_back(accu.back() + *iter);    }    int sumRange(int i, int j) {        return accu[j + 1] - accu[i];    }private:    vector<int> accu;    vector<int>::iterator iter;};