题目

给定一个整数数组  nums，求出数组从索引 i 到 j（i ≤ j）范围内元素的总和，包含 i、j 两点。

实现 NumArray 类：

NumArray(int[] nums) 使用数组 nums 初始化对象 int sumRange(int i, int j) 返回数组 nums 从索引 i 到 j（i ≤ j）范围内元素的总和，包含 i、j 两点（也就是 sum(nums[i], nums[i + 1], ... , nums[j])）  

示例：

输入：
["NumArray", "sumRange", "sumRange", "sumRange"]
[[[-2, 0, 3, -5, 2, -1]], [0, 2], [2, 5], [0, 5]]
输出：
[null, 1, -1, -3]

解释：
NumArray numArray = new NumArray([-2, 0, 3, -5, 2, -1]);
numArray.sumRange(0, 2); // return 1 ((-2) + 0 + 3)
numArray.sumRange(2, 5); // return -1 (3 + (-5) + 2 + (-1)) 
numArray.sumRange(0, 5); // return -3 ((-2) + 0 + 3 + (-5) + 2 + (-1))
``` 

提示：

0 <= nums.length <= 104
-105 <= nums[i] <= 105
0 <= i <= j < nums.length
最多调用 104 次 sumRange 方法

## 解题思路

class NumArray: # numArr = list() # snumArr = [0] def init(self, nums: List[int]): # self.numArr = list(nums) self.snumArr = [0] # 前缀和算法 for index,val in enumerate(nums): self.snumArr.append(self.snumArr[index]+val) # print(self.snumArr)

def sumRange(self, left: int, right: int) -> int:
    # 粗暴解法
    # print(left,right)
    # print(self.numArr[left:right])
    # return sum(self.numArr[left:right+1])
    # 前缀和算法
    return self.snumArr[right+1] - self.snumArr[left]

Your NumArray object will be instantiated and called as such:

obj = NumArray(nums)

param_1 = obj.sumRange(left,right)

numArray = NumArray([-2, 0, 3, -5, 2, -1])

print(numArray.sumRange(0, 2))

print(numArray.sumRange(2, 5))

print(numArray.sumRange(0, 5))

numArray = NumArray([-1]) print(numArray.sumRange(0, 0))