1. Two Sum

题目链接：https://leetcode.com/problems/two-sum/description/
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].

最简单的暴力破解：
def twoSum(self, nums, target):    72ms
    for i in range(len(nums)):
        for j in range(i+1,len(nums)):
            if nums[i]+nums[j]==target:
                return [i,j]

def twoSum(self, nums, target):    68ms   
    for i in nums:
        index0=nums.index(i)
        nums[index0]='*'
        if target-i in nums:
            index1=nums.index(target-i)
            return [index0,index1]

def twoSum(self, nums, target):##nums有重复值  36ms
    for i in nums:
        index0=nums.index(i)
        nums[index0]='*'
        if target-i in nums:
            index1=nums.index(target-i)
            return [index0,index1]

def twoSum1(self, nums, target):##不适合有重复值   36ms
    nums_={}
    for i in nums:
        nums_[target-i]=nums.index(i)
    for j in nums:
        nums_.pop(target-j)
        if j in nums_.keys():
            index1=nums.index(j)
            index2=nums_[j]
            return [index1,index2]
如果是递增有序的数组，可以设置两个指针，分别从左端右端往中间移动
然后如果和大于target,右指针左移一位，如果小于则左指针右移一位，待和为target或者左右指针相同返回。