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:

You may assume that the array does not change.

There are many calls to sumRange function

【思路】挨个遍历会超时，应该计算到每个元素之前的值。动态规划的思想。

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

}

int sumRange(int i, int j) {
if(i==0)
return subSum[j];
if(i<=0||i>subSum.size()||j>subSum.size()||i>j)
return 0;
return subSum[j] - subSum[i-1];
}
private:
vector<int> subSum;
};

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