leetcode lintcode python
目录
1.Two Sum(Hash Table)
Given an array of integers,return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution,and you may not use the same element twice.
Given nums=[2,7,11,15],target=[9],
Because nums[0]+nums[1]=2+7=9,
return [0,1]hash表，通过判断target-当前的数的值是否在hash表中
def twoSum(nums,target):
hash={}
for i in range(len(nums)):
if target-nums[i] not in hash:
hash[nums[i]]=i
else:
return [dict[target-nums[i]],i]
def twoSum(nums,target):
result=[]
for i in range(len(nums)):
for j in range(i+1,len(nums)):
if nums[i]+nums[j]==target:
result.append(i)
result.append(j)
return result
可以看出 hash表可以提高速度
9.回文数
输入：x = 121 输出：true
输入：x = -121 输出：false 解释：从左向右读, 为 -121 。 从右向左读, 为 121- 。因此它不是一个回文数。
输入：x = -121 输出：false 解释：从左向右读, 为 -121 。 从右向左读, 为 121- 。因此它不是一个回文数。
class Solution:
def isPalindrome(self, x: int) -> bool:
if x<0 or(x%10==0 and x!=0):
return False
rever=0
while(x>rever):
rever=rever*10+x%10
#取商
x=x//10
return x==rever or x==rever//1014.最长公共前缀
编写一个函数来查找字符串数组中的最长公共前缀
输入：strs = ["flower","flow","flight"] 输出："fl"
输入：strs = ["dog","racecar","car"] 输出："" 解释：输入不存在公共前缀。
class Solution:
def longestCommonPrefix(self, strs: List[str]) -> str:
res=""
for tmp in zip(*strs):
s=set(tmp)
if(len(s)==1):
res+=tmp[0]
else:
break
return res
15.三数之和
「不重复」的本质是什么？我们保持三重循环的大框架不变，只需要保证：
第二重循环枚举到的元素不小于当前第一重循环枚举到的元素；
第三重循环枚举到的元素不小于当前第二重循环枚举到的元素。
20.Valid Parentheses有效的括号(stack)
Given a string containing just the characters '(' ')' '{' '}' '[' ']', determine if the input string is valid.
An input string is valid if:
1.Open brackets must be closed by the same type of brackets.
2.Open brackets must be closed in the correct order.
Note that an empty string is also considered valid.
"([)]"
false#1.当遍历完字符串后栈不为空，则说明有多余的左括号，
#2.当遍历过程中遇到右括号时栈为空，则说明右括号多余,
#3.当遍历过程中遇到右括号时，栈顶元素与之不对应，则说明括号的类型没有对应上
class Solution:
def isValid(self, s: str) -> bool:
hash={')':'(','}':'{',']':'['}
stack=[]
for char in s:
if char not in hash:
stack.append(char)
else:
#如果栈为空，即情况2 or 栈顶元素不对应 即类型3
if not stack or stack[-1]!= hash[char]:
return False
stack.pop()
#如果栈为空 return true ,or return false
return not stack
#2
class Solution:
def isValid(self, s: str) -> bool:
dict={'(':')','{':'}','[':']'}
record=[]
for c in s:
if c=='(':
record.append('(')
elif c=='[':
record.append('[')
elif c=='{':
record.append('{')
else:
if record:
out=record.pop()
if dict[out]!= c:
return False
else:
return False
if record==[]:
return True
else :
return False
22.Generate Parentheses(括号生成)
Given n pairs of parentheses,write a function to generate all combinations of well-formed parentheses.
输入：n = 3 输出：["((()))","(()())","(())()","()(())","()()()"]
输入：n = 1 输出：["()"]
35.搜索插入位置
class Solution:
def searchInsert(self, nums: List[int], target: int) -> int:
left,right=0,len(nums)-1
while(left<=right):
middle=(right+left)//2
if nums[middle]>target:
right=middle-1
elif nums[middle]
42.接雨水
class Solution:
def trap(self, height: List[int]) -> int:
if not height:
return 0