LeetCode 303. Range Sum Query - Immutable

求数组nums[i,j]的和

思路：另开一sum数组，sum[i]为nums[0,i]的和，所以nums[i,j] = sum[j] - sum[i-1]

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

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

 

但是输入为[]时出现Runtime Error，将sum数组的长度+1，AC。

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

    int sumRange(int i, int j) {
        return sums[j+1]-sums[i];
    }
};