383. Ransom Note [easy] (Python)

题目链接

https://leetcode.com/problems/ransom-note/

题目原文

> Given
 an 
arbitrary
 ransom
 note
 string 
and 
another 
string 
containing 
letters from
 all 
the 
magazines,
 write 
a 
function 
that 
will 
return 
true 
if 
the 
ransom 
 note 
can 
be 
constructed 
from 
the 
magazines ; 
otherwise, 
it 
will 
return 
false. 



> Each 
letter
 in
 the
 magazine 
string 
can
 only 
be
 used 
once
 in
 your 
ransom
 note.

Note:
You may assume that both strings contain only lowercase letters.

canConstruct("a", "b") -> falsecanConstruct("aa", "ab") -> falsecanConstruct("aa", "aab") -> true

题目翻译

给定一个勒索信字符串，和一个包含所有杂志里字符的字符串，写一个函数，当该信可以被所有杂志里字符构成时返回true，否则返回false。杂志字符串里每个字符只能使用一次。
注意：假定两个字符串都只包含小写字符。

思路方法

思路一

计算两个字符串每个字母出现的次数再比较即可。可以用数组或字典保存每个字母出现的次数，对于两个字符串分别用加法和减法统计字母出现次数。下面的代码用了数组。

代码

class Solution(object):    def canConstruct(self, ransomNote, magazine):        """        :type ransomNote: str        :type magazine: str        :rtype: bool        """        if len(ransomNote) > len(magazine):            return False        letters = [0] * 26        for c in magazine:            letters[ord(c) - 97] += 1        for c in ransomNote:            letters[ord(c) - 97] -= 1            if letters[ord(c) - 97] < 0:                return False        return True

思路二

与思路一相同，不过用的字典。

代码

class Solution(object):    def canConstruct(self, ransomNote, magazine):        """        :type ransomNote: str        :type magazine: str        :rtype: bool        """        if len(ransomNote) > len(magazine):            return False        letters = {}        for c in magazine:            letters[c] = letters[c] + 1 if c in letters else 1        for c in ransomNote:            if c not in letters:                return False            letters[c] -= 1            if letters[c] < 0:                return False        return True

PS: 新手刷LeetCode，新手写博客，写错了或者写的不清楚还请帮忙指出，谢谢！
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