1. Two Sum

problem

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.

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

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
UPDATE (2016/2/13):
The return format had been changed to zero-based indices. Please read the above updated description carefully.

solution

• 大致思路

1. target减去每个元素=y
2. 判断y是否在序列里，是的话返回两个元素下标

• 特殊情况有两点

1. 例如[5,3,7],10 不能5+5=10;这一点用倒序pop移除元素
2. 例如[5,3,5],10 两个5的index要区分；这一点，第二个元素采用长度-倒序元素位置=正序元素位置（刚好上一点pop已经对list倒序了）

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        listNum = []
        mid = nums[::-1]
        for index,x in enumerate(nums):
            y = target - x
            mid.pop()
            if y in mid:
               listNum.append(index)
               listNum.append(len(nums)-1-mid.index(y))
               return listNum

• 遍历O(n)内嵌套了in O(n)；

discuss

class Solution(object):
    def twoSum(self, nums, target):
        if len(nums) <= 1:
            return False
        buff_dict = {}
        for i in range(len(nums)):
            if nums[i] in buff_dict:
                return [buff_dict[nums[i]], i]
            else:
                buff_dict[target - nums[i]] = i

• 大方向类似，逻辑更清晰一点，dict取元素的平均复杂度是O(1)

• else里面往字典里储存的是{差值：当前元素index}; 当某元素等于差值时，返回[之前储存的index,某元素index]