[Leetcode] 303. 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:

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

思路：

记录一个和数组sums，其中sums[j]表示从nums[0]到nums[j]的和。sums的状态转移方程是：sums[j] = sums[j - 1] + nums[j]，可以通过对数组一遍扫描完成；计算i到j之间的和可以通过sums[i, j] = sums[j] - sums[i-1]获得。

代码：

class NumArray {public:    NumArray(vector<int> nums) {        // the time complexity is O(n)        if (nums.size() > 0) {            sums.push_back(nums[0]);            for (int i = 1; i < nums.size(); ++i) {                sums.push_back(sums[i - 1] + nums[i]);            }        }    }        int sumRange(int i, int j) {        // the time complexity is O(1)        if (i == 0) {            return sums[j];        }        else {            return sums[j] - sums[i - 1];        }    }    private:    vector<int> sums;};/** * Your NumArray object will be instantiated and called as such: * NumArray obj = new NumArray(nums); * int param_1 = obj.sumRange(i,j); */