leetcode 303. Range Sum Query - Immutable

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

解题思路：
用一个数组dp保存nums的和，dp[i]表示nums[0...i]的和
求sumRange(i,j)即为dp[j]-dp[i-1];注意边界
*/

#include <iostream>
#include <string>
#include <vector>

using namespace std;

class NumArray {

private: vector<int> dp;
public:
    NumArray(vector<int> &nums) 
    {
        if (nums.size() == 0)
            return;
        dp.push_back(nums[0]);
        for (int i = 1; i < nums.size(); ++i)
        {
            dp.push_back(dp[i - 1] + nums[i]);
            //cout << dp[i] << endl;
        }
    }

    int sumRange(int i, int j) 
    {
        if (i == 0)
            return dp[j];
        return dp[j] - dp[i-1];
    }
};


// Your NumArray object will be instantiated and called as such:
// NumArray numArray(nums);
// numArray.sumRange(0, 1);
// numArray.sumRange(1, 2);

void test()
{
    vector<int> nums{ -2, 0, 3, -5, 2, -1 };
    NumArray numArray(nums);
    cout << numArray.sumRange(0, 1) << endl;
    cout << numArray.sumRange(1, 2) << endl;

}

int main()
{
    test();

    return 0;
}