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Commit042241d

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Update to 019
Update to 019
1 parent90ad977 commit042241d

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3 files changed

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3 files changed

+172
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#!usr/bin/env python3
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# -*- coding:utf-8 -*-
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'''
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Given a digit string, return all possible letter combinations that the number could represent.
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A mapping of digit to letters (just like on the telephone buttons) is given below.
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Input:Digit string "23"
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Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
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Note:
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Although the above answer is in lexicographical order, your answer could be in any order you want.
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'''
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classSolution:
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digit2letters= {
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'2':"abc",
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'3':"def",
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'4':"ghi",
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'5':"jkl",
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'6':"mno",
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'7':"pqrs",
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'8':"tuv",
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'9':"wxyz",
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}
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defletterCombinations(self,digits):
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"""
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:type digits: str
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:rtype: List[str]
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"""
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ifnotdigits:
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return []
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result= []
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self.dfs(digits,"",result)
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returnresult
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defdfs(self,digits,current,result):
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ifnotdigits:
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result.append(current)
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return
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forcinself.digit2letters[digits[0]]:
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self.dfs(digits[1:],current+c,result)
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if__name__=="__main__":
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assertSolution().letterCombinations("23")== ["ad","ae","af","bd","be","bf","cd","ce","cf"]
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‎Python3/018_4Sum.py

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#!usr/bin/env python3
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# -*- coding:utf-8 -*-
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'''
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Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.
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Note:
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Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
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The solution set must not contain duplicate quadruplets.
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For example, given array S = {1 0 -1 0 -2 2}, and target = 0.
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A solution set is:
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(-1, 0, 0, 1)
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(-2, -1, 1, 2)
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(-2, 0, 0, 2)
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'''
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classSolution(object):
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deffourSum(self,nums,target):
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"""
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:type nums: List[int]
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:type target: int
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:rtype: List[List[int]]
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"""
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iflen(nums)<4:
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return []
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result=set()
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sumsIndexes= {}
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foriinrange(len(nums)):
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forjinrange(i+1,len(nums)):
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ifnums[i]+nums[j]insumsIndexes:
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sumsIndexes[nums[i]+nums[j]].append((i,j))
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else:
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sumsIndexes[nums[i]+nums[j]]= [(i,j)]
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foriinrange(len(nums)):
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forjinrange(i+1,len(nums)):
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sumNeeded=target- (nums[i]+nums[j])
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ifsumNeededinsumsIndexes:
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forindexinsumsIndexes[sumNeeded]:
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ifindex[0]>j:
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result.add(tuple(sorted([nums[i],nums[j],nums[index[0]],nums[index[1]]])))
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result= [list(l)forlinresult]
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returnresult
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if__name__=="__main__":
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assertSolution().fourSum([1,0,-1,0,-2,2],0)== [[-1,0,0,1], [-2,0,0,2], [-2,-1,1,2]]
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#!usr/bin/env python3
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# -*- coding:utf-8 -*-
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'''
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Given a linked list, remove the nth node from the end of list and return its head.
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For example,
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Given linked list: 1->2->3->4->5, and n = 2.
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After removing the second node from the end, the linked list becomes 1->2->3->5.
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Note:
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Given n will always be valid.
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Try to do this in one pass.
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'''
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# Definition for singly-linked list.
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classListNode(object):
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def__init__(self,x):
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self.val=x
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self.next=None
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# Define this to check if it works well
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defmyPrint(self):
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print(self.val)
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ifself.next:
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self.next.myPrint()
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classSolution(object):
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defremoveNthFromEnd(self,head,n):
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"""
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:type head: ListNode
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:type n: int
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:rtype: ListNode
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"""
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ifnothead:
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returnhead
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point=ListNode(-1)
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point.next=head
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prev=point
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cur=point
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whileprevandn>=0:
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prev=prev.next
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n-=1
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whileprev:
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prev=prev.next
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cur=cur.next
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cur.next=cur.next.next
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returnpoint.next
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if__name__=="__main__":
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n5=ListNode(5)
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n4=ListNode(4)
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n3=ListNode(3)
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n2=ListNode(2)
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n1=ListNode(1)
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n1.next=n2
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n2.next=n3
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n3.next=n4
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n4.next=n5
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result=Solution().removeNthFromEnd(n1,5)
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result.myPrint()
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