# 「力扣」第 344 题:反转字符串(简单)
# 题目描述
编写一个函数,其作用是将输入的字符串反转过来。输入字符串以字符数组 char[]
的形式给出。
不要给另外的数组分配额外的空间,你必须原地修改输入数组、使用 O(1) 的额外空间解决这一问题。
示例 1:
输入:["h","e","l","l","o"]
输出:["o","l","l","e","h"]
示例 2:
输入:["H","a","n","n","a","h"]
输出:["h","a","n","n","a","H"]
提示:
1 <= s.length <= 10^5
s[i]
都是 ASCII (opens new window) 码表中的可打印字符
# 方法一:使用 Java 语言提供的反转 API 完成
参考代码 1:
public class Solution {
public String reverseString(String s) {
StringBuilder reverse = new StringBuilder();
for (int i = s.length()-1; i >=0 ; i--) {
reverse.append(s.charAt(i));
}
return reverse.toString();
}
// Given s = "hello", return "olleh".
public static void main(String[] args) {
String s = "hello";
Solution solution = new Solution();
String reverseString = solution.reverseString(s);
System.out.println(reverseString);
}
}
# 方法二:使用指针对撞
参考代码 2:
Java 代码:
public class Solution {
public String reverseString(String s) {
char[] cArray = s.toCharArray();
int i = 0;
int j = cArray.length - 1;
while (i < j) {
swap(cArray, i, j);
i++;
j--;
}
return new String(cArray);
}
private void swap(char[] s, int index1, int index2) {
if (index1 == index2) return;
char temp = s[index1];
s[index1] = s[index2];
s[index2] = temp;
}
public static void main(String[] args) {
Solution solution = new Solution();
String result = solution.reverseString("hello world");
System.out.println(result);
}
}
Python 代码:
class Solution(object):
def reverseString(self, s):
"""
:type s: str
:rtype: str
"""
if len(s) < 2:
return s
left = 0
right = len(s) - 1
l = list(s)
# 重合在一个就没有交换的必要了,因此是 left < right
while left < right:
l[left], l[right] = l[right], l[left]
left += 1
right -= 1
return ''.join(l)
作者:liweiwei1419 链接:https://suanfa8.com/two-pointers/solutions-1/0344-reverse-string 来源:算法吧 著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。