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range_sum_query_immutable.py
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range_sum_query_immutable.py
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"""
303. 区域和检索 - 数组不可变
设计 数组 前缀和
简单
给定一个整数数组 nums,处理以下类型的多个查询:
计算索引 left 和 right (包含 left 和 right)之间的 nums 元素的 和 ,其中 left <= right
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 中索引 left 和 right 之间的元素的 总和 ,包含 left 和 right 两点(也就是 nums[left] + nums[left + 1] + ... + nums[right] )
示例 1:
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:
1 <= nums.length <= 10^4
-10^5 <= nums[i] <= 10^5
0 <= i <= j < nums.length
最多调用 10^4 次 sumRange 方法
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/range-sum-query-immutable
"""
from typing import List
class NumArray:
def __init__(self, nums: List[int]):
sum_list = [nums[0]]
for i in range(1, len(nums)):
sum_list.append(sum_list[i-1] + nums[i])
self.sum_list = sum_list
def sumRange(self, left: int, right: int) -> int:
diff = 0
if left > 0:
diff = self.sum_list[left-1]
return self.sum_list[right] - diff
if __name__ == '__main__':
solution = NumArray([-2, 0, 3, -5, 2, -1])
assert solution.sumRange(0, 2) == 1
assert solution.sumRange(2, 5) == -1
assert solution.sumRange(0, 5) == -3