303. Range Sum Query - Immutable

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) -> 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.

Analysis:
注意note中最后一句manycalls， 所以如果在sumRange 函数中执行for循环那就太费时间了。
还有在做这道题的时候对于for循环注意到了一个问题：for循环先执行判断再执行循环体最后累加

还有就是vector::size 是unsign int型，所以可以试一下：

unsigned int k=0;cout<<k-1;

有时候这一点不注意会出错。
Source Code（C++）:

#include <iostream>#include <vector>#include <cmath>using namespace std;class NumArray {public:    vector<int> nums_sum;    NumArray(vector<int> &nums) {        int temp_sum=0;        for(int i=0; i<nums.size(); i++) {            temp_sum +=nums.at(i);            nums_sum.push_back(temp_sum);        }    }    int sumRange(int i, int j) {        if (i<0 || j>=nums_sum.size() || i>j){            return 0;        }        if (i==0){            return nums_sum.at(j);        }        else {            return nums_sum.at(j)-nums_sum.at(i-1);        }    }};int main() {    vector<int> nums;    nums.push_back(-2);    nums.push_back(0);    nums.push_back(3);    nums.push_back(-5);    nums.push_back(2);    nums.push_back(-1);    NumArray numArray(nums);    cout << numArray.sumRange(0, 2);    cout << numArray.sumRange(0, 5);    return 0;}