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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))
