LeetCode 303. Range Sum Query - Immutable

题目：

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.

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

思路：

　　自底向上得到所有sumRange(0, i)的解，保存在public成员solution中； Θ(n)

　　sumRange(i, j) = sumRange(0, j) - sumRange(0, i - 1);

代码：C++

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

int sumRange(int i, int j) {
if (i < 0 || j >= num.size())
return 0;
if (i == 0)
return solution[j];
else
return solution[j] - solution[i - 1];
}

vector<int> solution;

private:
vector<int> num;

};