找出字符串差异与勒索信构建-优快云博客

本文介绍两道编程题的解决方案：一是通过对比两个字符串找出被添加的字符；二是判断是否能从一组杂志中构建出勒索信。两题均采用计数法解决，通过对字符出现次数的统计来实现目标。

389.Find the Difference

Description：

Given two strings s and t which consist of only lowercase letters.

String t is generated by random shuffling string s and then add one more letter at a random position.

Find the letter that was added in t.

Example:

Input:
s = "abcd"
t = "abcde"

Output:
e

Explanation:
'e' is the letter that was added.

分析：

   首先要确定不能被这一个样例哄骗了。s是随机产生的，t在s的基础上加上一个字母，字母的位置也是随机的。

My C++ code：

class Solution {
public:
    char findTheDifference(string s, string t){
        int m = s.length() ;
        int n = t.length() ;
        char res ;
        vector<int> letter(26, 0) ;
        for(int i = 0 ; i < m ; i ++)
        {
            letter[s[i] - 'a'] ++ ; 
        }
        for(int i = 0 ; i < n ; i ++)
        {
            letter[t[i] - 'a'] -- ;
        }
        for(int i = 0 ; i < 26 ; i ++)
        {
            if(letter[i] < 0)
            {
                res = i + 'a';
                break ;
            }
        }
        return res ;
    }
};

383.Ransom Note

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

分析：

   套路和上一道题目是一样的，注意要分清楚magazines和randsomNote。

My C++ code：

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        vector<int> letter(26 , 0) ;
        int m = magazine.size() ;
        int n = ransomNote.size() ;
        for(int i = 0 ; i < n ; i ++)
        {
            letter[ransomNote[i] - 'a'] ++ ;
        }
        for(int i = 0 ; i < m ; i ++)
        {
            letter[magazine[i] - 'a'] -- ;
        }
        for(int i = 0 ; i < 26 ; i ++)
        {
            if(letter[i] > 0)
            {
                return false ;
            }
        }
        return true ;
    }
};