Complex Number Multiplication
Given two strings representing two complex numbers.
You need to return a string representing their multiplication. Note i2 = -1 according to the definition.
Example 1:
Input: "1+1i", "1+1i" Output: "0+2i" Explanation: (1 + i) * (1 + i) = 1 + i2 + 2 * i = 2i, and you need convert it to the form of 0+2i.
Example 2:
Input: "1+-1i", "1+-1i" Output: "0+-2i" Explanation: (1 - i) * (1 - i) = 1 + i2 - 2 * i = -2i, and you need convert it to the form of 0+-2i.
Note:
- The input strings will not have extra blank.
- The input strings will be given in the form of a+bi, where the integer a and b will both belong to the range of [-100, 100]. And the output should be also in this form.
class Solution(object):
def complexNumberMultiply(self, a, b):
"""
:type a: str
:type b: str
:rtype: str
"""
a=a.rstrip("i").split("+")
b=b.rstrip("i").split("+")
aa=int(a[0])
ab=int(a[1])
ba=int(b[0])
bb=int(b[1])
shishuPart=aa*ba-ab*bb
xushuPart=aa*bb+ab*ba
return str(shishuPart)+"+"+str(xushuPart)+"i"Nth Digit
Find the nth digit of the infinite integer sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...
Note:
n is positive and will fit within the range of a 32-bit signed integer (n < 231).
Example 1:
Input: 3 Output: 3
Example 2:
Input: 11 Output: 0 Explanation: The 11th digit of the sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ... is a 0, which is part of the number 10.
class Solution(object):
def findNthDigit(self, n):
"""
:type n: int
:rtype: int
"""
strN=str(n)
length=len(strN)
if length==1:
return n
n=n-9 #1位数的总共有9个，减去这9个数
for i in range(2,length+1):
if n>i*(10**i-10**(i-1)):
n-= i*(10**i-10**(i-1))
else:
if n % i == 0:
numberIndex=int(n/i)
else:numberIndex=int(n/i)+1
charIndex=n%i
number=10**(i-1)+numberIndex-1
if charIndex==0:
return int(str(number)[charIndex-1])
return int(str(number)[charIndex-1])Pascal's Triangle
Given numRows, generate the first numRows of Pascal's triangle.
For example, given numRows = 5,
Return
[
[1],
[1,1],
[1,2,1],
[1,3,3,1],
[1,4,6,4,1]
]
class Solution(object):
def generate(self, numRows):
"""
:type numRows: int
:rtype: List[List[int]]
"""
if numRows==0:
return []
if numRows==1:
return [[1]]
if numRows==2:
return [[1],[1,1]]
result=[[1],[1,1]]
for i in range(numRows-2):
length=len(result)
lastElement=result[length-1]
toBeAppended=[1]
for i in range(length-1):
toBeAppended.append(lastElement[i]+lastElement[i+1])
toBeAppended.append(1)
result.append(toBeAppended)
return resultBase 7
Given an integer, return its base 7 string representation.
Example 1:
Input: 100 Output: "202"
Example 2:
Input: -7 Output: "-10"
Note: The input will be in range of [-1e7, 1e7].
class Solution(object):
def convertToBase7(self, num):
"""
:type num: int
:rtype: str
"""
result=""
isNegtive=False
if num < 0:
isNegtive=True
num=-num
Count=1
while 7**Count <= num:
Count+=1
for i in range(Count):
if num>=7**(Count-i-1):
result+=str(int(num/7**(Count-i-1)))
num=num%7**(Count-i-1)
else:
result+="0"
if isNegtive:
return "-"+result
else:
return resultBinary Watch
A binary watch has 4 LEDs on the top which represent the hours (0-11), and the 6 LEDs on the bottom represent the minutes (0-59).
Each LED represents a zero or one, with the least significant bit on the right.
For example, the above binary watch reads "3:25".
Given a non-negative integer n which represents the number of LEDs that are currently on, return all possible times the watch could represent.
Example:
Input: n = 1
Return: ["1:00", "2:00", "4:00", "8:00", "0:01", "0:02", "0:04", "0:08", "0:16", "0:32"]
Note:
- The order of output does not matter.
- The hour must not contain a leading zero, for example "01:00" is not valid, it should be "1:00".
- The minute must be consist of two digits and may contain a leading zero, for example "10:2" is not valid, it should be "10:02".
import itertools
class Solution(object):
def readBinaryWatch(self, num):
"""
:type num: int
:rtype: List[str]
这是一个组合的问题
"""
result=[]
combinations=list(itertools.combinations(range(0,10),num))
for i in combinations:
strTime=list("0000000000")
for j in i:
strTime[j]=1
hour=0
minute=0
for j in range(6):
if strTime[j]==1:
minute+=2**j
if minute>=60:
continue
for j in [6,7,8,9]:
if strTime[j]==1:
hour+=2**(j-6)
if hour>=12:
continue
if minute<10:
minute="0"+str(minute)
result.append(str(hour)+":"+str(minute))
return resultLongest Substring Without Repeating Characters
Given a string, find the length of the longest substring without repeating characters.
Examples:
Given "abcabcbb", the answer is "abc", which the length is 3.
Given "bbbbb", the answer is "b", with the length of 1.
Given "pwwkew", the answer is "wke", with the length of 3. Note that the answer must be a substring, "pwke" is a subsequence and not a substring.
class Solution(object):
def lengthOfLongestSubstring(self, s):
"""
:type s: str
:rtype: int
一个一个往前 , 如果遇到相同的，如abcb，那么从第一个字符截断，变成cb，再往下找。
"""
if s=="":
return 0
length=len(s)
if length==1:
return 1
maxString=1
tmpS=""
for i in s:
tmpS=tmpS+i
setS=set(tmpS)
if len(setS)<len(tmpS): #如果这时候遇到重复的字符串了。
repeatCharIndex=tmpS.index(tmpS[len(tmpS)-1])
tmpS=tmpS[repeatCharIndex+1:]
setS=set(tmpS)
else:
maxString=max(maxString,len(tmpS))
return maxString