Python 刷Leetcode题库，顺带学英语单词（3）

Palindrome Number

Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward. [#9]

   Example 1:

   Input: 121

   Output: true

   Example 2:

   Input: -121

   Output: false

   Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.

   Example 3:

   Input: 10

   Output: false

   Explanation: Reads 01 from right to left. Therefore it is not a palindrome.

Follow up:

Coud you solve it without converting the integer to a string?

题意：判断一个整数是否为回文数？ 后续：能否不转成字符串就解决问题。

>>> result = lambda n:n==int(str(n)[::-1]) if n>=0 else False
>>> result(121)
True
>>> result(-121)
False
>>> result(10)
False
>>> 

如不允许转字符串，可以用%和//逐位取出数字再作判断。

>>> def isPalindrome(number):
  if number<0:
    return False
  res = []
  while number != 0:
    res.append(number%10)
    number //= 10
  return res==res[::-1]
>>> isPalindrome(121)
True
>>> isPalindrome(-121)
False
>>> isPalindrome(10)
False
>>> isPalindrome(0)
True
>>> 

Container With Most Water

Given n non-negative integers a1, a2, ..., an , where each represents a point at coordinate (i, ai).n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water. [#11]

Note: You may not slant the container and n is at least 2.

The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.

   Example:

   Input: [1,8,6,2,5,4,8,3,7]

   Output: 49

题意：找出序列中的元素所能围成矩形的最大面积值。

再次用到“子序列推导式”，只是本题不需要单个元素的子序列，很简单只要把后一个range()的第一参数i+1改成i+2：

>>> s=[1,2,3]
>>> [s[i:j] for i in range(len(s)) for j in range(i+1,len(s)+1)]
[[1], [1, 2], [1, 2, 3], [2], [2, 3], [3]]
>>> [s[i:j] for i in range(len(s)) for j in range(i+2,len(s)+1)]
[[1, 2], [1, 2, 3], [2, 3]]
>>> 

到得的子序列只要把首尾两元素的小者乘以（子序列长度-1）就是面积，其面积的最大值就是最终答案：

>>> maxArea = lambda s:max([min(t[0],t[-1])*(len(t)-1) for t in [s[i:j] for i in range(len(s)) for j in range(i+2,len(s)+1)]])
>>> L = [1,8,6,2,5,4,8,3,7]
>>> maxArea(L)
49
>>> 

Roman to Integer

Roman numerals are represented by seven different symbols: I , V , X , L , C , D and M .

Symbol Value

   I          1

   V         5

   X         10

   L          50

   C         100

   D         500

   M        1000

For example, two is written as II in Roman numeral, just two one's added together. Twelve is written as, XII , which is simply X + II . The number twenty seven is written as XXVII , which is XX + V + II .

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII . Instead, the number four is written as IV . Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX .

There are six instances where subtraction is used:

I can be placed before V (5) and X (10) to make 4 and 9.

X can be placed before L (50) and C (100) to make 40 and 90.

C can be placed before D (500) and M (1000) to make 400 and 900.

Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.   [#13]

题意：把罗马数字的字符串转换成一个十进制整数。

>>> def roman2int(n):
  last = total = 0
  for i in range(len(n)):
    c = n[len(n)-(i+1):len(n)-i]
    num = c in 'IXCM' and 10**'IXCM'.index(c) or c in 'VLD' and 5*10**'VLD'.index(c)
    total += num*(-1 if num < last else 1)
    last = num
  return total
>>> roman2int('III')
3
>>> roman2int('IV')
4
>>> roman2int('IX')
9
>>> roman2int('LVIII')
58
>>> roman2int('MCMXCIV')
1994
>>> roman2int('MMXXI')
2021
>>> 

相关单词