leetcode-DP-   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:

1. You may assume that the array does not change.
2. There are many calls to sumRange function.

DP解法，先计算0-第i项的和，最终结果就是0-j的和减去0-i-1的和

class NumArray {
public:
	vector<int> sums;
	int s;
	NumArray(  vector<int> nums)
	{
		s = nums.size();
		if (s <= 0)return ;
		sums.reserve(100);

		sums.push_back(nums[0]);

		for (int i = 1; i < nums.size(); i++)
		{

	 
			int a = sums[i - 1] + nums[i];

			sums.push_back(a);


		}

	}


	int sumRange(int i, int j) {
		if (s <= 0)return 0;

		if (i == 0)return sums[j];
		return sums[j]-sums[i-1];
	}
 


};