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.

动态规划.

class NumArray {
public:
NumArray(vector<int> &nums) {
if(!nums_.empty()) {
nums_.clear();
}
if(!sums_.empty()) {
sums_.clear();
}
if(nums.empty()) return;
for(int i = 0; i < nums.size(); i++) {
nums_.push_back(nums[i]);
// if(i == 0) {
//     sums_.push_back(nums_[i]);
// } else {
//     sums_.push_back(nums_[i] + sums_[i - 1]);
// }
}
sums_.push_back(nums_[0]);
for(int i = 1; i < nums.size(); i++) {
sums_.push_back(nums_[i] + sums_[i - 1]);
}
}

int sumRange(int i, int j) {
int sum = 0;
if(j < i) std::swap(i, j);
// for(int idx = i; idx <= j; idx++) {
//     sum += nums_[idx];
// }
return sums_[j] - sums_[i] + nums_[i];
return sum;
}
private:
vector<int> nums_;
vector<int> sums_;
};

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