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LeetCode-1--两数之和(Two-Sum)

2019-03-24
leacoder

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两数之和

给定一个整数数组 nums 和一个目标值 target,请你在该数组中找出和为目标值的那 两个 整数,并返回他们的数组下标。

你可以假设每种输入只会对应一个答案。但是,你不能重复利用这个数组中同样的元素。

示例:

给定 nums = [2, 7, 11, 15], target = 9

因为 nums[0] + nums[1] = 2 + 7 = 9
所以返回 [0, 1]

Python3实现

暴力循环

# @author:leacoder
# @des:   暴力循环  两数之和  O(N*N)
class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        for index1,num1 in enumerate(nums):  #两层循环
            for index2,num2 in enumerate(nums[index1+1:]):
                if num1 + num2 == target:
                    return [index1,index1+index2+1]
        return []

和固定利用差值diff

# @author:leacoder
# @des:   和固定利用差值diff 两数之和 O(N)
class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        for i in range(len(nums)-1):#循环  O(N)
            diff = target - nums[i]
            if diff in nums[i+1:]: #O(1)
                return [i,nums[i+1:].index(diff)+i+1]
        return []

GitHub链接: https://github.com/lichangke/LeetCode

知乎个人首页: https://www.zhihu.com/people/lichangke/

简书个人首页: https://www.jianshu.com/u/3e95c7555dc7

个人Blog: https://lichangke.github.io/

欢迎大家来一起交流学习


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