345 / 344 Reverse Vowels of a String / Reverse String

345 Reverse Vowels of a String

Total Accepted: 537  Total Submissions: 1488  Difficulty: Easy

Write a function that takes a string as input and reverse only the vowels of a string.

Example 1:
Given s = "hello", return "holle".

Example 2:
Given s = "leetcode", return "leotcede".

分析：DONE

典型的双指针问题：从前从后分别寻找（利用set加快查找速度）元音的位置，然后交换，注意边界条件即可！

class Solution {
public:
    void swapchar(string &str,int pos1,int pos2)  
    {  
        char tmpch=str[pos2];  
        str[pos2]=str[pos1];  
        str[pos1]=tmpch;  
    }  
    string reverseVowels(string s) {
        char base[10]={'a','e','i','o','u','A','E','I','O','U'};
        set seting(base,base+10);
        int startpos=0,endpos=s.size()-1;
        while(startpos < endpos )
        {
            while( startpos < endpos && seting.find(s[startpos]) == seting.end())//从左往右、新的、未被交换过的元音位置
                startpos++;
            while( startpos < endpos && seting.find(s[endpos]) == seting.end())
                endpos--;
            if(startpos < endpos)    
                swapchar(s,startpos++,endpos--);    
        }
        return s;
    }
};

344 Reverse String

Total Accepted: 598  Total Submissions: 936  Difficulty: Easy

Write a function that takes a string as input and returns the string reversed.

Example:
Given s = "hello", return "olleh".

分析：DONE

字符串反转！恐怕此题会成为leetcode历史上最简单的题目！

时间复杂度o(n)

class Solution {
public:
    void swapchar(string &str,int pos1,int pos2)
    {
        char tmpch=str[pos2];
        str[pos2]=str[pos1];
        str[pos1]=tmpch;
    }
    string reverseString(string s) {
        int halfsize=s.size()/2;
        for(int i=0;i < halfsize;i++)
            swapchar(s,i,s.size()-i-1);
        return s;
    }
};

或者：

class Solution {
public:
    string reverseString(string s) {
        reverse(s.begin(), s.end());
        return s;
    }
};

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原文地址：http://blog.csdn.net/ebowtang/article/details/51228095

原作者博客：http://blog.csdn.net/ebowtang

本博客LeetCode题解索引：http://blog.csdn.net/ebowtang/article/details/50668895