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];
}
};