LeetCode 15. 3Sum--Java，Python解法

题目地址：

Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

The solution set must not contain duplicate triplets.

Example:

Given array nums = [-1, 0, 1, 2, -1, -4],A solution set is:
[[-1, 0, 1],[-1, -1, 2]
]

谷歌翻译的题目：
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。

注意：

解决方案集不得包含重复的三元组。

我的思路是2重循环，算出a+b的所有可能，然后判断-(a+b)是否在列表里，时间复杂度是O(N^2)
Python解法：

class Solution:def threeSum(self, nums = [-4,-2,1,-5,-4,-4,4,-2,0,4,0,-2,3,1,-5,0]):nums.sort()length=len(nums)temp=[]temp4=set()a={}b={}for i in nums:if i not in a:a[i]=1else:a[i]+=1i=0while i<length:j=i+1while j<length:passif j>=i+2 and nums[j]==nums[j-1]:if nums[j]==nums[i]:i=j-2breakj+=1continueif nums[i] not in b:b[nums[i]]=set()if nums[j] in b[nums[i]]:j+=1continueelse:b[nums[i]].add(nums[j])temp2=-(nums[i]+nums[j])if temp2 in a :count=1if nums[i]==temp2:count+=1if nums[j]==temp2:count+=1if a[temp2]>=count:temp3=[nums[i],nums[j],temp2]temp3.sort()if str(temp3) not in temp4:temp4.add(str(temp3))temp.append([nums[i],nums[j],temp2])j+=1i+=1                       for i in temp:i.sort()temp.sort()return temp

改进后的办法是两边向中间逼近：

class Solution:def threeSum(self, nums = [-4,-2,1,-5,-4,-4,4,-2,0,4,0,-2,3,1,-5,0]):res = []nums.sort()length = len(nums)for i in range(0,length-2): if nums[i]>0: break if i>0 and nums[i]==nums[i-1]: continue l, r = i+1, length-1 while l<r:total = nums[i]+nums[l]+nums[r]if total<0:l+=1elif total>0: r-=1else: res.append([nums[i], nums[l], nums[r]])while l<r and nums[l]==nums[l+1]:l+=1while l<r and nums[r]==nums[r-1]: r-=1l+=1r-=1return res

Java解法如下：

class Solution {public List<List<Integer>> threeSum(int[] nums) {List<List<Integer>> res = new ArrayList<>();Arrays.sort(nums);for (int i = 0; i + 2 < nums.length; i++) {if (i > 0 && nums[i] == nums[i - 1]) {              // skip same resultcontinue;}int j = i + 1, k = nums.length - 1;  int target = -nums[i];while (j < k) {if (nums[j] + nums[k] == target) {res.add(Arrays.asList(nums[i], nums[j], nums[k]));j++;k--;while (j < k && nums[j] == nums[j - 1]) j++;  // skip same resultwhile (j < k && nums[k] == nums[k + 1]) k--;  // skip same result} else if (nums[j] + nums[k] > target) {k--;} else {j++;}}}return res;
}
}

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