LeetCode : 345. Reverse Vowels of a String

 

这个题目本身很简单，主要是性能的问题，用string自带的函数find_first_of 和find_last_of，运行了72ms

class Solution {
public:
    string reverseVowels(string s) {
        string temp(s);
        bool loop = true;
        std::size_t posS = std::string::npos;
        std::size_t posE = std::string::npos;
        int offset=0;

        while(loop){
            posS = temp.find_first_of("aeiouAEIOU");
            posE = temp.find_last_of("aeiouAEIOU");
            if(posS<posE){
                char tempC = s[posS+offset];
                s[posS+offset] = s[posE+offset];
                s[posE+offset] = tempC;
                offset = offset + posS + 1;
                temp = temp.substr(posS+1,posE-posS-1);
            }else{
                loop = false;
            }
        }
        return s;
    }
};

用处理数组的方式从两边向中间找用了20ms.

class Solution {
public:
    string reverseVowels(string s) {
        int len = s.length();
        int ii=0,jj=len-1;
        bool loop = true,found = false;
        
        while(loop){
            for(int i=ii;i<len;i++){
                if(s[i]=='a'||s[i]=='e'||s[i]=='i'||s[i]=='o'||s[i]=='u'
                   ||s[i]=='A'||s[i]=='E'||s[i]=='I'||s[i]=='O'||s[i]=='U'){
                       ii = i;
                       found = true;
                       break;
                   }
                   
            }
            for(int j=jj;j>=0;j--){
                if(s[j]=='a'||s[j]=='e'||s[j]=='i'||s[j]=='o'||s[j]=='u'
                   ||s[j]=='A'||s[j]=='E'||s[j]=='I'||s[j]=='O'||s[j]=='U'){
                       jj = j;
                       found = true;
                       break;
                   }
            }
            if(ii<jj && found){
                char tempC = s[ii];
                s[ii] = s[jj];
                s[jj] = tempC;
            }else{
                break;
            }
            ii++;
            jj--;
        }
        return s;
    }
};