1. Two Sum

题目

给你一个列表nums和一个数字target，如果列表里面有两个数之和等于target，则返回两个数字的下坐标，这两个数不能为同一个数。(假设总会有这两个数)

python

时间复杂度O(n^2)

# -*- coding:utf-8 -*-

class Solution(object): # O(n^2)
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        for num in nums:
        	i = nums.index(num)
        	j = [j for j,v in enumerate(nums) if v==(target-num) and j!=i]
        	if j:
        		return [i, j[0]]

s = Solution()
print(s.twoSum([0, 4, 3, 0], 0))

时间复杂度O(n)

# -*- coding:utf-8 -*-

class Solution(object): # O(n)
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        map = {}
        for index, num in  enumerate(nums):
            if num in map:
                return [map[num], index]
            else:
                map[target-num] = index

                
s = Solution()
print(s.twoSum([0, 4, 3, 0], 0))