LeetCode-1 - 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].

Solution1

很容易想到循环遍历列表，复杂度为O(n2)

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        for i in range(len(nums) - 1):
            for j in range(i + 1, len(nums)):
                if nums[i] + nums[j] == target:                    
                    return [i, j]

Solution2

利用字典来实现，算法复杂度为O(n)

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        map = {}
        for i, a in enumerate(nums):
            if target - a in map:
                return [map[a], i]
            map[target - a] = i

        return -1, -1

反思

1. 对于solution过程中需要用到下标的题目，enumerate通常比较实用
2. map的查找效率为O(1)
3. 两层循环可以利用map做cache将其简化，然后利用map的高效查找降低算法复杂度
4. x + y = z的问题，变成了z - y = x的问题，其中z, y 为已知，x的查找转化成map的查找