LeetCode 383:Ransom Note (c++)
2017-06-22 15:03
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一:题目
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.
二:解题思路
一开始看题,并没有看懂题目的意思,查阅了网友们的博客,该题可以理解为:
给定一个字符串A,然后再给定一个字符串B。要求如果A字符串能够从B字符串的字符中构造,那么返回true,否则返回false
即,在字符串A中出现的字符个数要小于字符串B相应的字符个数
则只需统计字符串A中,出现了哪些字符,出现多少次,字符串B中出现哪些字符,出现多少次,然后两者进行比较
三:代码实现
class Solution {
public:
bool canConstruct(string ransomNote, string magazine) {
int i,j;
int valueOfRansomNote[26]={0};
int valueOfMagazine[26]={0};
//统计ransomNote中的26个字母出现的个数
for(i=0;i<ransomNote.length();i++)
valueOfRansomNote[ransomNote[i]-'a']++;
//统计magazine中的26个字母出现的个数
for(j=0;j<magazine.length();j++)
valueOfMagazine[magazine[j]-'a']++;
//两个字符串26个字母出现次数比较
for(i=0;i<26;i++)
if(valueOfRansomNote[i]>valueOfMagazine[i])
return false;
return true;
}
};
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
二:解题思路
一开始看题,并没有看懂题目的意思,查阅了网友们的博客,该题可以理解为:
给定一个字符串A,然后再给定一个字符串B。要求如果A字符串能够从B字符串的字符中构造,那么返回true,否则返回false
即,在字符串A中出现的字符个数要小于字符串B相应的字符个数
则只需统计字符串A中,出现了哪些字符,出现多少次,字符串B中出现哪些字符,出现多少次,然后两者进行比较
三:代码实现
class Solution {
public:
bool canConstruct(string ransomNote, string magazine) {
int i,j;
int valueOfRansomNote[26]={0};
int valueOfMagazine[26]={0};
//统计ransomNote中的26个字母出现的个数
for(i=0;i<ransomNote.length();i++)
valueOfRansomNote[ransomNote[i]-'a']++;
//统计magazine中的26个字母出现的个数
for(j=0;j<magazine.length();j++)
valueOfMagazine[magazine[j]-'a']++;
//两个字符串26个字母出现次数比较
for(i=0;i<26;i++)
if(valueOfRansomNote[i]>valueOfMagazine[i])
return false;
return true;
}
};
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