【leetcode】345. Reverse Vowels of a String

一、题目描述

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".

题目解读：将字符串s中的元音逆置

思路：遍历一遍字符串，用map存储元音和位置，然后设置两个迭代器分别从头和从尾遍历，修改元音位置

c++代码（64ms，4.59%）

class Solution {
public:
string reverseVowels(string s) {
map<int, char>vowel;
int count = 0;
for(int i=0; i<s.size(); 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'){
vowel[i] = s[i];
count ++;
}
}
if(count){
map<int, char>::iterator it1;
map<int, char>::iterator it2;
it1 = vowel.begin();
it2 = vowel.end();
it2--;
for(int i= 0;i<count; i++){
s[it1->first] = it2->second;
it1++;
it2--;
}
}
return s;
}
};

其他代码

class Solution {
public:
string reverseVowels(string s) {
auto p1 = s.begin(), p2 = s.end() - 1;
string vowels = "aeiouAEIOU";
while(p1 < p2) {
while((vowels.find(*p1) == string::npos) && (p1 < p2)) p1++;
while((vowels.find(*p2) == string::npos) && (p1 < p2)) p2--;
if(p1 < p2) swap(*p1, *p2);
p1++;
p2--;
}
return s;
}
};

代码2 (12ms, 73.65%)

class Solution {
public:
string reverseVowels(string s) {
int dict[256] = {0};
dict['a'] = 1, dict['A'] = 1;
dict['e'] = 1, dict['E'] = 1;
dict['i'] = 1, dict['I'] = 1;
dict['o'] = 1, dict['O'] = 1;
dict['u'] = 1, dict['U'] = 1;
int start = 0, end = (int)s.size() - 1;
while(start < end){
while(start < end && dict[s[start]] == 0) start++;
while(start < end && dict[s[end]] == 0) end--;
swap(s[start],s[end]);
start++;end--;
}
return s;
}
};