3Sum

Given an array S of n integers, are there elements a, b, c in S 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.

For example, given array S = [-1, 0, 1, 2, -1, -4], A solution set is: [ [-1, 0, 1], [-1, -1, 2] ]

 

分析：

方法一：暴力解法时间复杂度为，肯定超时

方法二：先将数组排序，然后依次遍历，每次固定一个值，使用两个指针一前一后遍历固定值之后的元素，注意避免重复元素，

　　代码一：

class Solution(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        nums.sort()
        res = []
        for i in range(len(nums)-2):
            l = i+1
            r = len(nums)-1
            while l < r:
                while l < r and nums[i]+nums[l]+nums[r] == 0:
                    res_tmp = [nums[i],nums[l],nums[r]]
                    if res_tmp not in res:
                        res.append(res_tmp)
                    while l < r and nums[l+1] == nums[l]:
                        l += 1
                    while l < r and nums[r-1] == nums[r]:
                        r -= 1
                    l += 1
                    r -= 1
                while l < r and nums[i]+nums[l]+nums[r] > 0:
                    r -= 1
                while l < r and nums[i]+nums[l]+nums[r] < 0:
                    l += 1
        return res

此代码最终也是超时

　　代码二：

class Solution(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        size = len(nums)
        ans = []
        if size <= 2:
            return ans
        nums.sort()
        i = 0
        while i < size -2:
            tmp = 0 - nums[i]
            j = i + 1
            k = size -1
            while j < k:
                if nums[j] + nums[k] < tmp:
                    j += 1
                elif nums[j] + nums[k] > tmp:
                    k -= 1
                else:
                    ans.append([nums[i],nums[j],nums[k]])
                    j += 1
                    k -= 1
                    while j < k:
                        if nums[j] != nums[j - 1]:
                            break
                        if nums[k] != nums[k + 1]:
                            break
                        j += 1
                        k -= 1
            i += 1
            while i < size - 2:
                if nums[i] != nums[i - 1]:
                    break
                i += 1
        return ans

此代码可以AC，注意和代码一做对比。