[Leetcode] 1.Two Sum @python

题目

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

暴力解法

两层循环遍历所有可能的给定数组中两个数的组合，时间复杂度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 + 1,j + 1]        return []

高效解法

对数组进行排序，设置头指针和尾指针，分别从两端向中间移动，直到找到两个指针指向的数字和与target相等。这里需要注意，因为要求返回在原数组中的下标，所以需要对原数组进行拷贝，以查找得到的两个数字在原数组中的下标.时间复杂度O(N)

class Solution(object):    def twoSum(self, nums, target):        """        :type nums: List[int]        :type target: int        :rtype: List[int]        """        old_nums = list(nums) # copy original array        nums.sort()        left = 0        right = len(nums) - 1        while left < right:            if nums[left] + nums[right] == target:                break            elif nums[left] + nums[right] < target:                left += 1            else:                right -= 1        ans = []        for i in range(len(old_nums)):            if  old_nums[i] == nums[left]                  or old_nums[i] == nums[right]:                ans.append(i + 1)                if len(ans) == 2:                    break        return ans