LeetCode 303. 区域和检索 - 数组不可变

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))