题目:
Write a function that reverses a string. The input string is given as an array of characters char[].
Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
You may assume all the characters consist of printable ascii characters.
Example 1:
Input: ["h","e","l","l","o"]
Output: ["o","l","l","e","h"]
Example 2:
Input: ["H","a","n","n","a","h"]
Output: ["h","a","n","n","a","H"]
本题使用对撞指针解决此问题,速度很快
public static void reverseString(char[] s) {
if (s == null || s.length == 0)
return;
int l = 0;
int r = s.length - 1;
while (l <= r){
swap(s, l++, r--);
}
}
public static void swap(char[]s, int i, int j){
char temp = s[i];
s[i] = s[j];
s[j] = temp;
}
public static void main(String[] args) {
char[] s = {'h', 'e', 'l', 'l', 'o'};
reverseString(s);
System.out.println(Arrays.toString(s));
}
}
博客给出一道反转字符串的LeetCode题目,要求不额外分配数组空间,在原数组上以O(1)额外内存完成反转。给出示例输入输出,还指出使用对撞指针可快速解决该问题。
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