My Python solution (accept)

  • 0
    class Solution:
        # @return a list of lists of length 3, [[val1,val2,val3]]
        def threeSum(self, num):
            returnlists = []
            #let num be sorted, ....
            lasta = None
            #let a = num[i], this solution is to find (b + c) = -a
            for i in range(len(num)):
                #use lasta to avoid redundant a
                if num[i] == lasta:
                #use lastc to avoid redundant c
                lastc = None
                lasta = num[i]
                #this is -a
                nega = 0 - num[i]
                #make a set to store the b, then to find c
                newSet = set()
                #use range(i,..) to avoid (a, b)/(b, a) redundancies.
                for j in range(i, len(num)):
                    #skip self
                    if j != i:
                        #check if  -a - c is in set
                        theval = nega - num[j]
                        #not in and no repeating, get (a, b, c).
                        if theval in newSet and lastc != num[j]:
                            thislist = [theval, num[i], num[j]]
                            lastc = num[j]
                        #put b in set
            return returnlists

Log in to reply

Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.