先看一篇大神寫的文章：為什么要刷題
http://selfboot.cn/2016/07/24/leetcode_guide_why/

內(nèi)容：
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].

大值意思是存在一個(gè)數(shù)值列表，給定一個(gè)數(shù)值，判斷列表內(nèi)是否存在兩個(gè)數(shù)是否相加等于給定的數(shù)值，若存在則返回對(duì)應(yīng)的索引值。

參考答案：

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        for i, num in enumerate(nums):
            other = target - num 
            try:
                position = nums.index(other)
                if position != i:
                    return sorted([i,position])
            except ValueError:
                pass 

思路：

使用要判斷的數(shù)與列表中的數(shù)相減，判斷獲得的差值是否存在列表中，如果存在則返回對(duì)應(yīng)的序列號(hào)

知識(shí)點(diǎn)：

1. 返回列表的索引 enumerate()
2. 由值獲得列表對(duì)應(yīng)的索引 index()
3. 列表排序 sorted