[剑指Offer]41 和为S的两个数字 VS 和为S的连续正数序列

Leetcode T1 Two Sum

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example

Given nums = [2, 7, 11, 15], target = 9.

Because nums[0] + nums[1] = 2 + 7 = 9,

return [0, 1].

题目：给一个序列和一个目标值，返回相加等于目标值的序列中元素的索引。

思路：遍历序列，用i记录序列第一个值的索引，值记录为val。用diff = target - val（# val1 + val2 = target），查看diff索引。当diff ！= i 时，返回[i, diff的索引]。

代码

class Solution:

    def twoSum(self, nums: List[int], target: int) -> List[int]:

        for i in range(len(nums)):

            val = nums[i]

            diff = target - val

            if diff in nums and i != nums.index(diff):

                return [i, nums.index(diff)]

提交结果

Runtime: 1204 ms, faster than 23.29% of Python3 online submissions for Two Sum.

Memory Usage: 13.8 MB, less than 69.77% of Python3 online submissions for Two Sum.

巴特西

[剑指Offer]41.和为S的两个数字 VS 和为S的连续正数序列

[剑指Offer]41 和为S的两个数字 VS 和为S的连续正数序列

Leetcode T1 Two Sum

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