鉴于对算法的感觉太少,广度和深度都不够,故开启刷leetcode,在此记录
Easy 篇
1. Two sum
# Given nums = [2, 7, 11, 15], target = 9,
# Because nums[0] + nums[1] = 2 + 7 = 9,
# return [0, 1].
# nums = [1,2,3,4,5]
# print({i:n for i,n in enumerate(nums)})
class Solution:
def twoSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
for i, n in enumerate(nums):
for j, m in enumerate(nums):
if n + m == target and i != j:
return [i,j]
def twoSum2(self, nums, target):
dic = dict()
for index,value in enumerate(nums):
sub = target - value # 3:0 2:1
if sub in dic:
return [dic[sub],index]
else:
dic[value] = index
print(Solution().twoSum2(nums=[3,2,4],target=6))
# [1, 2]
第一种方式比较直观,双重遍历数组解决问题
第二种采用由和变差的方式,采用字典存储value与index.当当前dic中的key与之前的sub相等,说明target对应的index已找到
7. Reverse Integer
class Solution:
def reverse(self, x):
"""
:type x: int
:rtype: int
"""
y = x if x > 0 else -x
z = 0
while y > 0 :
z = z * 10 + y % 10
y //= 10
z = z if x > 0 else -z
if z > 0xffffffff or z < -0xffffffff:
return 0
else:
return z
print(Solution().reverse(123))
通过拆分每一个数字的各个位,再进行逆序叠加
注意: 在Python3中, a / b,执行真除法,无论操作数的类型怎样,结果都会返回浮点型.
a // b, 执行Floor除法,返回截除余数的整数部分,若操作数中有浮点型,则结果为浮点型
此题中,若使用 a / b,会出现y=0依然能进入到while循环中,此处为Python的Bug.
9. Palindrome Number
# Determine whether an integer is a palindrome. Do this without extra space.
#Example 1234321
class Solution:
def isPalindrome(self, x):
"""
:type x: int
:rtype: bool
"""
if x < 0:
return False
des = []
while x > 0:
des.append(x % 10)
x //= 10
dic = {index:value for index,value in enumerate(des)}
index = 0
sub = 0
dic_len = len(dic) - 1
for x in range(0,dic_len):
sub = dic[index] - dic[dic_len - index]
if sub != 0:
return False;
return True;
def isPalindrome2(self,x):
if x < 0:
return False
render = 1
while x / render >= 10:
render *= 10
# render 为 10000
while x:
left = x / render
right = x % 10
if left != right:
return False
x = (x % render) // 10
render /= 100
return True
print(Solution().isPalindrome(12321))
方法一将int数转化为以位数为key的字典,通过循环比较首位和对应的尾位,是否相同来判断是否为回文数
方法二直接将int数的首位和尾位提取出来进行比较
20. Valid Parentheses
# Given a string containing just the characters '(', ')', '{', '}', '[' and ']',
#determine if the input string is valid.
# The brackets must close in the correct order,
# "()" and "()[]{}" are all valid but "(]" and "([)]" are not.
class Solution(object):
def isValid(self, s):
"""
:type s: str
:rtype: bool
"""
# ord('a') --> 97
# chr(97) --> a
# numList = list(map(lambda x:ord(x), s))
m = {'}':'{', ']':'[', ')':'('}
tempList = []
for e in s:
if tempList and (e in m and tempList[0] == m[e]):
tempList.pop()
else:
tempList.append(e)
return not tempList
print(Solution().isValid("[]{}()"))
比较巧妙的点1. 既然要判断[ ] { } ( )成对存在,则将其转为字典做判断
- 数组做中间变量存储,如果至少不存在一对成对,则数组不为空
由于是判断 [ 和 ], 所以 ] 为 key, [ 为value