LeetCode网 - 0001：Two Sum

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

You can return the answer in any order.

 

Example 1:

Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].

Example 2:

Input: nums = [3,2,4], target = 6
Output: [1,2]

Example 3:

Input: nums = [3,3], target = 6
Output: [0,1]

 

Constraints:

• 2 <= nums.length <= 104
• -109 <= nums[i] <= 109
• -109 <= target <= 109
• Only one valid answer exists.

 

Follow-up: Can you come up with an algorithm that is less than O(n2) time complexity?

 

思路一：

要找「两数」之和，所以枚举所有「两数」的组合，从而找到对应的解

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

时间复杂度：O(n^2)

空间复杂度：O(1)

 

 

思路二：

 执一搜一，将「两数」分解成：一个已知和一个未知。从而转化成：在一个「无序」数组中搜索一个给定数字。只搜一次的话简单遍历即可，此时我们只使用O(1)的辅助空间，而现在的问题是「在一个不断插入新数字的<序列>中，如何快速查找到某一个数字？」这里的<序列>也可以叫做<容器>，即使用辅助存储空间的情况。为了追求速度，我们选择hash表作为辅助存储结构。

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        index = {}
        for i, x in enumerate(nums):
            if target - x in index:
                return [index[target - x], i]
            else:
                index[x] = i

时间复杂度：O(n)

空间复杂度：O(n)