LeetCode —-Two Sum

题目

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].

解决

首先想到的是采用双层循环，解决这个问题。代码如下：

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

运行结果虽然没有错误，但是提交代码时出现TLM,超时了。看来服务器限制算法空间复杂度了，不支持O(n^2)

所以下边考虑修改空间复杂度，反复考虑后采用反证法解决复杂度问题。哈哈哈哈（反证法）。复杂度降低为O(n),完美提交

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