303. Range Sum Query

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) -> 1sumRange(2, 5) -> -1sumRange(0, 5) -> -3

Note:

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

题解：

这题比较容易。一种方法是对于sumRange（i,j）可以遍历从i到j的数据然后进行相加；另一种可行的算法是（以下代码是指这一方法）可以用从0-j的数据和减去0-i的数据和。

具体代码如下：

class NumArray {public:    NumArray(vector<int> nums) {        int len = nums.size();         sum = new int[len];         int k = 0;         for (int i = 0; i < len; i ++) {            k += nums[i];             sum[i] = k;                     }    }        int sumRange(int i, int j) {        if (i == 0) {             return sum[j];         }         if (j == 0) {            return sum[0];        }        return sum[j] - sum[i - 1];     }private:     int *sum; };/** * Your NumArray object will be instantiated and called as such: * NumArray obj = new NumArray(nums); * int param_1 = obj.sumRange(i,j); */

end!