leetcode-mid-array-31 three sum-NO

my code:    time limited

def threeSum(nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """     
        def twoSum(a,b,c):
            if a + b + c == 0:
                return True
            else:
                return False
            
        res = []
        for i in range(len(nums)-2):
            a = nums[i]
            for j in range(i+1,len(nums)-1):
                b = nums[j]
                k = j +1
                #print(i,j,k,a,b,nums[k])
                while k < len(nums):
                    print(a,b,nums[k])
                    if twoSum(a,b,nums[k]) :
                        print('....',a,b,nums[k])
                        temp = [a,b,nums[k]]
                        temp.sort()
                        print('...temp',temp)
                        if temp not in res:
                            res.append(temp)
                        print('....res',res)
                    k += 1   
        return res

思路：

1、时间复杂度n*n

排序 ( WT1：为什么要排序？）-》构建字典记录value：次数-》两层循环，其中不断不判断第三个数是不是也等于两层循环的值（用到dic的记录次数）

class Solution(object):
    
    '''
    O(n^2)时间复杂度
    
    '''
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        result = []
        num_cnt_dict = {}
        
        for num in nums:
            num_cnt_dict[num] = num_cnt_dict.get(num, 0) + 1
            
        if 0 in num_cnt_dict and num_cnt_dict[0] >= 3:
            result.append([0, 0, 0])
        
        nums = sorted(list(num_cnt_dict.keys()))
        
        for i, num in enumerate(nums):
            for j in nums[i+1:]:
                if num * 2 + j == 0 and num_cnt_dict[num] >=2:
                    result.append([num, num, j])
                if j * 2 + num == 0 and num_cnt_dict[j] >= 2:
                    result.append([j, j, num])
                
                diff = 0 - num - j
                if diff > j and diff in num_cnt_dict:
                    result.append([num, j, diff])
        return result

注意，以下的方式就会超时，可能是因为list和dict.keys()的检索查找方式不同

                if diff > b and diff in nums:  #要注意>b，才能不和前面的if重合
                    res.append([a,b,diff])
        return res

2、n*m 其中n+m=len(nums)

构建dict -》 不排序，将dic的keys分为正数和负数-》两层for循环

注意下面这一行：是因为filter得到的是有序递增数列！！！！！！！所以当diff小于a时，说明当i=diff时候，由于不满足这个条件，还没有把[a,b,diff]放入最后结果

if diff<a or diff>b:
　　 res.append([a,b,diff])

class Solution(object):
    
    '''
    O(n^2)时间复杂度
    
    '''
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        res = []
        dic = {}
        for item in nums:
            dic[item] = dic.get(item,0) + 1   
        if 0 in dic and dic[0] >= 3: 
            res.append([0,0,0]) #为什么这个要单独列出来 -》因为除了这种情况，for循环中不会再出现三个数相等且和为0的情况
        #nums = sorted(list(dic.keys())) 
        neg = list(filter(lambda x:x<0,dic.keys()))
        pos = list(filter(lambda x:x>=0,dic.keys()))
        for a in neg:
            for b in pos:#为什么j的有边界不是len(nums)-1 -》因为可能会存在第三个数巧合和nums[j]相等
                diff  = 0 - a - b
                if diff in dic:
                    if diff in (a,b) and dic[diff] >= 2: 
                        res.append([a,b,diff])
                    if diff<a or diff>b:
                        res.append([a,b,diff])
        return res