383 Ransom Note 赎金信
Description:
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") -> false
canConstruct("aa", "ab") -> false
canConstruct("aa", "aab") -> true
题目描述:
给定一个赎金信 (ransom) 字符串和一个杂志(magazine)字符串，判断第一个字符串ransom能不能由第二个字符串magazines里面的字符构成。如果可以构成，返回 true ；否则返回 false。
(题目说明：为了不暴露赎金信字迹，要从杂志上搜索各个需要的字母，组成单词来表达意思。)
注意：
你可以假设两个字符串均只含有小写字母。
canConstruct("a", "b") -> false
canConstruct("aa", "ab") -> false
canConstruct("aa", "aab") -> true
思路:
参考LeetCode #242 Valid Anagram 有效的字母异位词
可以建立一个 26位的数组用来记录 ransomNote出现的字符及次数
时间复杂度O(n), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
bool canConstruct(string ransomNote, string magazine)
{
vector<int> count(26, 0);
for (int i = 0; i < magazine.size(); i++) ++count[magazine[i] - 'a'];
for (int i = 0; i < ransomNote.size(); i++)
{
count[ransomNote[i] - 'a']--;
if (count[ransomNote[i] - 'a'] < 0) return false;
}
return true;
}
};
Java:
class Solution {
public boolean canConstruct(String ransomNote, String magazine) {
int count[] = new int[26];
for (int i = 0; i < magazine.length(); i++) count[magazine.charAt(i) - 'a']++;
for (int i = 0; i < ransomNote.length(); i++) {
count[ransomNote.charAt(i) - 'a']--;
if (count[ransomNote.charAt(i) - 'a'] < 0) return false;
}
return true;
}
}
Python:
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
return not collections.Counter(ransomNote) - collections.Counter(magazine)
