178 lines
4.2 KiB
Markdown
178 lines
4.2 KiB
Markdown
# 原创
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： leetcode15-三数之和
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# leetcode15-三数之和
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>
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题目详见：[https://leetcode-cn.com/problems/3sum/description/](https://leetcode-cn.com/problems/3sum/description/)
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>
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给定一个包含 n 个整数的数组 nums，判断 nums 中是否存在三个元素 a，b，c ，使得 a + b + c = 0 ？找出所有满足条件且不重复的三元组。
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注意：答案中不可以包含重复的三元组。
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```
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例如, 给定数组 nums = [-1, 0, 1, 2, -1, -4]，
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满足要求的三元组集合为：
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[
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[-1, 0, 1],
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[-1, -1, 2]
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]
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```
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这道题我遇到好几次了，B站后台开发工程师笔试题就考了这个。今天就来总结一下解决方案。<br/> 这道题并不难，正常人的第一反应肯定是三个循环，再简单不过了，可是时间复杂度是不是有点大。。。提交也不让通过是吧。<br/> 首先写好边界条件，数组小于三个元素就不满足条件：
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```
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result = []
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nums_len = len(nums)
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if nums_len < 3:
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return result
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```
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注意我们用的是Python语言，所以可以直接**先用内建函数`sort()`排序**，**方便操作。然后可以用一个for循环，两个指针遍历**，时间复杂度也就是
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o
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(
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n
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2
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)
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o(n^2)
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o(n2)，与之前的三循环减小了不少。了解了这些差不多就可以实现代码了。
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```
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class Solution(object):
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def threeSum(self, nums):
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"""
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:type nums: List[int]
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:rtype: List[List[int]]
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"""
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result = list()
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nums_len = len(nums)
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if nums_len < 3:
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return result
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l, r, dif = 0, 0, 0
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nums.sort()
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for i in range(nums_len - 2):
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if nums[i] > 0:
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break
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if i > 0 and nums[i - 1] == nums[i]:
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continue
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l = i + 1
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r = nums_len - 1
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dif = -nums[i]
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while l < r:
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if nums[l] + nums[r] == dif:
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result.append([nums[l], nums[r], nums[i]])
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# 这两个while完全可以不要
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while l < r and nums[l] == nums[l + 1]:
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l += 1
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while l < r and nums[r] == nums[r - 1]:
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r -= 1
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l += 1
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r -= 1
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elif nums[l] + nums[r] < dif:
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l += 1
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else:
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r -= 1
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return result
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```
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另外，leetcode16题也是差不多这个意思：
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>
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例如，给定数组 nums = [-1，2，1，-4], 和 target = 1.与 target 最接近的三个数的和为 2. (-1 + 2 + 1 = 2).
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思路一模一样，一个for，加上两个指针，时间复杂度为
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O
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(
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n
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2
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)
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O(n^2)
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O(n2)
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```
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class Solution(object):
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def threeSumClosest(self, nums, target):
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"""
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:type nums: List[int]
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:type target: int
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:rtype: int
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"""
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# n = len(nums)
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# if n<3:return None
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# res = sum(nums[0:3])
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# diff = abs(res - target)
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# for i in range(0,n-2):
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# for j in range(i+1,n-1):
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# for k in range(j+1,n):
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# tmp = nums[i]+nums[j]+nums[k]
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# if diff > abs(tmp - target):
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# diff = abs(tmp - target)
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# res = tmp
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# return res
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n = len(nums)
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if n<3:return None
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nums.sort()
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res = sum(nums[0:3])
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diff = abs(res - target)
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for i in xrange(len(nums)-2):
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low = i+1
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high = len(nums)-1
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while low<high:
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tmp = nums[i]+nums[low]+nums[high]
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if abs(res-target) > abs(tmp-target):res = tmp
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elif target < tmp: high -= 1
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else: low += 1
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return res
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```
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