Leetcode-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) -> 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.

以为很简单就做了，下面的方法超时啊啊啊，然后被超时虐了，因为很多调用。

class NumArray {
public:
    NumArray(vector<int> &nums) {
        num=nums;
    }

    int sumRange(int i, int j) {
        int sum=0;
        while(i<=j)
        {
            sum+=num[i];
        }
        return sum;
    }
private:
   vector<int> num;
};


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

只好换种思路，num[i]存的是nums的前i-1个元素之和，反正前j个元素和减掉前i-1个元素和等于i到j的和。

class NumArray {
public:
    NumArray(vector<int> &nums) {
        num.push_back(0);
        for(int i=1;i<=nums.size();i++)
        {
            num.push_back(num[i-1]+nums[i-1]);
        }
    }

    int sumRange(int i, int j) {
        return num[j+1]-num[i];
    }
private:
   vector<int> num;
};


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