15. 3Sum

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]
]

Solution1:(TLE)Brute Force

class Solution:
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        res = []
        for i in range(len(nums)):
            for j in range(i+1,len(nums)):
                for k in range(j+1,len(nums)):
                    if nums[i] + nums[j] + nums[k]==0:
                        res.append(tuple(sorted([nums[i],nums[j],nums[k]])))
        return [list(i) for i in set(res)]

267 / 313 test cases passed.

Solution2:（TLE） Better brute force

class Solution:
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        temp = set()
        nums.sort()
        # print(nums)
        for i in range(len(nums)-2):
            if nums[i]>0:  #多加了个剪枝
                break
            for j in range(i+1,len(nums)-1):
                if nums[i]+nums[j]>0:
                    break
                for k in range(j+1,len(nums)):
                    # print('i:',i,'j:',j,'k:',k)
                    # print(nums[i],' ',nums[j],' ',nums[k],' ',nums[i] + nums[j]+nums[k])
                    if nums[i]+nums[j]+nums[k]>0:
                        break
                    if nums[i] + nums[j]+nums[k]==0:
                        temp.add((nums[i],nums[j],nums[k]))
        return [list(i) for i in temp]

283 / 313 test cases passed.

Solution3:（TLE）

class Solution:
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        temp = set()
        nums.sort()
        # print(nums)
        for i in range(len(nums)-2):
            if nums[i]>0: 
                break
            for j in range(i+1,len(nums)-1):
                if nums[i]+nums[j]>0:
                    break
                for k in range(len(nums)-1,j,-1): #倒着搜索更快
                    # print('i:',i,'j:',j,'k:',k)
                    # print(nums[i],' ',nums[j],' ',nums[k],' ',nums[i] + nums[j]+nums[k])
                    if nums[i]+nums[j]+nums[k]<0:
                        break
                    if nums[i] + nums[j]+nums[k]==0:
                        temp.add((nums[i],nums[j],nums[k]))
        return [list(i) for i in temp]

311 / 313 test cases passed.

Solution4:（TLE）

class Solution:
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        temp = set()
        nums.sort()
        for i in range(len(nums)-2):
            if nums[i]>0:
                break
            left,right = i+1,len(nums)-1
            while left<right:
                if nums[i]+nums[left]+nums[right]>0:
                    right -= 1
                    continue
                if nums[i]+nums[left]+nums[right]<0:
                    left += 1
                    continue
                temp.add((nums[i],nums[left],nums[right]))
                left += 1
                right -= 1
        return [list(i) for i in temp]

312 / 313 test cases passed.
确定了第一项之后，对于后两项用有序序列求和的方式更快。

Solution5:

class Solution:
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        temp = set()
        nums.sort()
        if len(nums)>=3 and nums[0]==nums[-1]==0:  对于全0的数组单独判断
            return [[0,0,0]]
        for i in range(len(nums)-2):
            if nums[i]>0:
                break
            left,right = i+1,len(nums)-1
            while left<right:
                if nums[i]+nums[left]+nums[right]>0:
                    right -= 1
                    continue
                if nums[i]+nums[left]+nums[right]<0:
                    left += 1
                    continue
                temp.add((nums[i],nums[left],nums[right]))
                left += 1
                right -= 1
        return [list(i) for i in temp]