383.leetcode Ransom Note(easy)[字符统计]

来源：互联网发布：网络教育学校名单编辑：程序博客网时间：2024/05/15 10:39

 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

刚开始以为是子串匹配问题，后来发现是应该是字符统计问题，看ransomNote中需要的字符是否在magazine中出现足够的次数，

只要ransomNote出现的字符次数小于magazine中相同字符出现的次数即可。

class Solution {public:    //子串匹配问题    /*bool canConstruct(string ransomNote, string magazine) {        int n1 = ransomNote.size();        int n2 = magazine.size();        if(n1<=0||n2<=0) return false;        char start = ransomNote[0];        vector<int> pos;        for(int i=0;i<n2;i++)        {            if(magazine[i] == start)                pos.push_back(i);        }                for(int i=0;i<pos.size();i++)        {            int j = pos[i]+1;            int k = 1;            while(k<n1&&j<n2&&ransomNote[k] == magazine[j])            {                    ++k;                   ++j;            }            //cout<<k<<""<<j<<endl;            if(k==n1&&j<=n2)            {                return true;            }        }        return false;    }*/    //字符统计问题，看ransomNote中需要的字符是否否在magazine中出现足够的次数    bool canConstruct(string ransomNote, string magazine) {        int n1 = ransomNote.size();        int n2 = magazine.size();        if(n1 == 0 && n2 == 0) return true;        else if(n1 == 0) return true;        else if(n2 == 0) return false;        int status1[26];//ran统计        int status2[26];//magazine统计        memset(status1,0,sizeof(status1));        memset(status2,0,sizeof(status2));        for(int i=0;i<n1;i++)        {            status1[ransomNote[i]-'a']++;        }        for(int j=0;j<n2;j++)        {            status2[magazine[j]-'a']++;        }        for(int i=0;i<26;i++)        {            if(status1[i]>status2[i])               return false;        }        return true;    }};

0 0