15. 三数之和（3Sum）

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：
答案中不可以包含重复的三元组。
例如, 给定数组 nums = [-1, 0, 1, 2, -1, -4]，
满足要求的三元组集合为：
[
[-1, 0, 1],
[-1, -1, 2]
]

思考

思考的过程，我昨天晚上想了想，没想出来。然后，刚刚想的思路，大概还是有点像，无论对错，先将它写出来：

nums = [-1, 0, 1, 2, -1, -4]
答案：
-1 -1 2
-1 0 1
封装：
se = {-4: 1, -1: 2, 0: 1, 1: 1, 2: 1}
思考过程：
①一负数，两个正数：
-4 4 0 ✘
-4 3 1 ✘
-4 2 2 ✘
-1 1 0 ✔
②三个零
0 0 0 ✘
③一正数，两个负数：
1 -1 0 (重复，可以设置结果集合，然后去重)
2 -1 -1 ✔

class Solution(object):

    def twoSum(self, nums, target):
        result = []
        re = nums.copy()  # 浅拷贝，复制一个备份
        for i, j in enumerate(nums):
            r = target - j
            re.remove(nums[i]) #i是下标，j是值，按下标删除
            if r in re:
                if [j, r] not in result and [r, j] not in result:
                    result.append([j, r])

        return result

    def findZero(self, nums):
        """
        找到起始0元素，或者最小的非负数
        :param nums:
        :return: 下标位置
        """
        for i, j in enumerate(nums):
            if j>=0:
                return i

    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        #接收数据的列表
        result = []
        #①感觉封装有点多余，直接排序就可以了
        nums = sorted(nums)

        re = []
        #②处理数据，需要两个指针，一前一后扫描
        i, j = 0, len(nums)-1
        #循环的过程，应该是i扫描到末尾
        #j具有特殊性，如果i是负数，j从后往前；i是正数，j从前到后
        while i<len(nums):
            if nums[i]<0:
                #转化成求在[0-max]之间求两数之和==-nums[i],突然想到了用之前第一题中的two sum
                #但是这里可能有多个值，我们需要修改程序
                re = self.twoSum(nums[self.findZero(nums):], nums[i]*-1)
                #print(re) #[[1, 3], [2, 2]]
                re.append(nums[i])
                result.append(re)
                i+=1  
            elif nums[i]==0:
                #判断0的个数
                count = nums.count(0) // 3  #整除，得到[0,0,0]列表的个数
                # whlie count>0:
                if count>0:
                    result.append([0,0,0])
                    #count-=1，结果集无论几个[0,0,0]，都只加入一个
                #保证0处执行一次，我们后移i
                while i<len(nums) and nums[i]==0:
                    i+=1
            else:#  nums[i]>0
                # 同理，此时的范围是[min-0)
                re = self.twoSum(nums[:self.findZero(nums)], nums[i] * -1)
                re.append(nums[i])
                result.append(re)
                i+=1

        final = []
        #处理结果集#[[[1, 3], [2, 2], -4], [[0, 1], -1], [[0, 1], -1], [0, 0, 0], [[-1, -1], 2], [[-1, -1], 2], [3]]
        k = 0
        while k<len(result):
            i = result[k]
            if len(i)>1:
                j = 0
                while j<len(i)-1:
                    if isinstance(i[j], list):
                        ms = i[j].copy()  #[1, 3]
                        ms.append(i[len(i)-1])  #-4
                        if ms not in final:
                            final.append(ms)
                    else:
                        final.append(i)
                        j+=1
                    j+=1
            k+=1

        return final

结果：超时
提交记录
311 / 313 个通过测试用例

对于最后一个测试用例，我们这里演示一下：

import time
    start = time.clock()
    nums = 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    a = Solution().threeSum(nums)

    elapsed = (time.clock() - start)
    print("Time used:", elapsed)

然后，结果是：
Time used: 58.208637248206905

当然了，放入上面的目的只是为了说明我们的程序是可以运行的，但是不够简洁。
我们需要更高效率的程序代码。

不妨考虑考虑暴力法

先暴力解决出来，然后我们再考虑优化。

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：答案中不可以包含重复的三元组。

class Solution(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        re = []
        i = 0
        while i<len(nums):
            j = 0
            while j<len(nums):
                k = 0
                while k<len(nums):
                    if i!=j and j!=k :
                        if nums[i] + nums[j] == nums[k]*-1:
                            li = sorted([nums[i], nums[j], nums[k]])
                            if li not in re:
                                re.append(li)
                    k+=1
                j+=1
            i+=1

        return re

这个程序很容易理解，就是靠下标区别元素，以保证a，b，c三个数不是同一个位置，也就是不是一个数，两两不同。然后判断结果是否是我们期待的0。
但是：超出时间限制
也不难理解，因为我们的程序的时间复杂度是O(n^3)，效率不高。我们接下来在它的基础上优化。

class Solution(object):
    #数据封装
    def loadData(self, nums):
        data = dict()
        for i in nums:
            data[i] = data.get(i, 0) + 1
        return data

    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """

        re = []
        i = 0
        while i<len(nums):
            # 先将数据封装到字典中，方便计算数据的个数
            data = self.loadData(nums)
            j = 0
            data[nums[i]] -= 1
            while j<len(nums):
                data[nums[j]] -= 1
                if i!=j:
                    if data.get( -nums[i]-nums[j] , 0) > 0:
                        data[-nums[i]-nums[j]] -= 1
                        li = sorted([nums[i], nums[j], -nums[i]-nums[j]])
                        if li not in re:
                            re.append(li)
                j+=1
            i+=1
        return re

结果：
311 / 313 个通过测试用例
状态：超出时间限制

看来，任重而道远！！！

继续优化

百度了一个解题思路：

首先对数组从小到大排序，从一个数开始遍历，若该数大于0，后面的数不可能与其相加和为0，所以跳过；否则该数可能是满足要求的第一个数，这样可以转化为求后面数组中两数之和为该数的相反数的问题。定义两个指针一前一后，若找到两数之和满足条件则加入到解集中；若大于和则后指针向前移动，反之则前指针向后移动，直到前指针大于等于后指针。这样遍历第一个数直到数组的倒数第3位。注意再求和过程中首先判断该数字是否与前面数字重复，保证解集中没有重复解。

下面实现：

class Solution(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        re=[]
        #排序-从小到大
        nums.sort()
        #第一个数大于0,返回空
        if len(nums)<3 or nums[0] > 0 :
            return []
        #数据可能满足条件
        i=0
        while i<len(nums) - 2:
            #print(123)
            if nums[i]<=0:
                # 求后面数组中两数之和为该数的相反数的问题
                left, right = i + 1, len(nums) - 1
                while left < right:
                    s = nums[left] + nums[right] + nums[i]
                    # 满足条件,加入集合中
                    if s == 0:
                        li = sorted([nums[left], nums[right], nums[i]])
                        if li not in re:
                            re.append(li)
                        left += 1
                        right -= 1
                    elif s > 0:
                        right -= 1
                    else:
                        left += 1
            i += 1

        return re

但是和上面一样还是超时！！！

然后又找了找思路：

正确思路：
需要解决的问题是：如何把三重遍历变为双重遍历
解决思路也是重点：排序+双指针
1 先对nums进行排序，利用排序和两个“指针”，将三重遍历变为双重遍历，思路类似快排
2 另外，一定要对数据进行去重，否则一些极端输入如[0,0,0,0,0,0,0,0,0,0]，速度会非常慢
具体思路：
1 对异常数据进行处理，如[1,2]和[0,0,0]
2 因为已经排过序了，所以如果当前的元素大于0，那么后面的元素一定也大于0，必不符合条件，所以直接return
3 对起点进行去重之后，对当前元素之后的元素进行遍历，设置两个指针left和right，分别从左右向中间移动，小则left右移，大则right左移。此处必须保证去重，每次进行判定的必须是“新”的组合。
下面的算法运行时间在55~59ms之间

class Solution(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        re=[]
        #排序-从小到大
        nums.sort()
        #第一个数大于0,返回空
        if len(nums)<3 or nums[0] > 0 :
            return []
        if len(nums)==3 and nums[0]+nums[1]+nums[2]==0:
                li = sorted([nums[0], nums[1], nums[2]])
                re.append(li)
                return re
        if len(set(nums))==1 and nums[0] == 0:
            return [[0,0,0]]

        #数据可能满足条件
        i=0
        while i<len(nums) - 2:
            if nums[i]>0:
                return re

            # 求后面数组中两数之和为该数的相反数的问题
            left, right = i + 1, len(nums) - 1
            while left < right:
                s = nums[left] + nums[right] + nums[i]
                # 满足条件,加入集合中
                if s == 0:
                    li = sorted([nums[left], nums[right], nums[i]])
                    if li not in re:
                        re.append(li)
                    #左右去重
                    while left<right and nums[left+1] == nums[i]:
                        left += 1
                    while left<right and nums[right-1] == nums[right]:
                        right -= 1
                    left += 1
                    right -= 1
                elif s > 0:
                    right -= 1
                else:
                    left += 1
            i += 1

        return re