LeetCode 303. Range Sum Query - Immutable

较简单的动态规划题目：LeetCode 303. Range Sum Query - Immutable

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.

其实最简单的思路就是将所要求的区域内的数字相加，不过这样频繁调用会使得复杂度较大。

采用动态规划的思路，只调用一次构造函数即可。这样一来sumRange函数的复杂度降为O(1)。

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