[一起来刷leetcode吧][33]--No.15 3Sum

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这是leetcode的第15题--3Sum

  题目

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]
]

  思路 最开始我想的就直接3重循环，再加判重的循环，暴力求解，当然超时了，要高于O(n³)。后来想到可以将正负数，0，分成三组来组合，然而最后两个数据过不了，在网上搜了一下，可以固定一个数，头尾双指针来移动，这是O(n²)。哎，折腾了一晚上，我好菜啊。 这是前几次的结果

  

show me the code

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

        rst = []
        zero = []   #zeros
        neg = []    #negative
        pos = []    #positive
        for i in (nums):
            if i < 0:
                neg.append(i)
            elif i > 0:
                pos.append(i)
            else:
                zero.append(i)
        if len(zero) > 2:
            rst.append([0,0,0])
        if neg == []  or  pos == []:
            return rst
        if zero != []:
            if len(neg) > len(pos):
                for i in pos:
                    if -i in neg:
                        rst.append([-i,0,i])
            else:
                for i in neg:
                    if -i in pos:
                        rst.append([i,0,-i])
        pos.sort()
        neg.sort()
        if len(pos) == 1 and len(neg) == 1:
            return rst
        elif len(pos) == 1 :
            tmp = len(neg) - 1
            while tmp > 0:
                sum = neg[tmp]   neg[tmp-1]
                if sum == - pos[0]:
                    rst.append([neg[tmp-1],neg[tmp],pos[0]])
                    break
                elif sum < - pos[0]:
                    break
                tmp -= 1
        elif len(neg) == 1:
            tmp = 0
            while tmp < len(pos) - 1 :
                sum = pos[tmp]   pos[tmp 1]
                if sum == - neg[0]:
                    rst.append([neg[0],pos[tmp],pos[tmp 1]])
                    break
                elif sum > - neg[0]:
                    break
                tmp -= 1
        sameI = []     #avoid test several same num
        for i in range(len(pos)-1):
            if i in sameI:
                continue
            sameI.append(i)
            sameJ=[]
            for j in range(i 1,len(pos)):
                if j in sameJ:
                    continue
                sameJ.append(j)
                sum = pos[i]   pos[j]
                for k in neg:
                    if   sum > -k:
                        break
                    elif sum == -k:
                        rst.append([k,pos[i],pos[j]])
        sameI = []
        for i in range(len(neg)-1):
            if i in sameI:
                continue
            sameI.append(i)
            sameJ=[]
            for j in range(i 1,len(neg)):
                if j in sameJ:
                    continue
                sameJ.append(j)
                sum = neg[i]   neg[j]
                for k in pos:
                    if   sum > -k:
                        break
                    elif sum == -k:
                        rst.append([neg[i],neg[j],k])
        fnl = []   
        for i in rst:
            if i not in fnl:
                fnl.append(i)
        return fnl