leetcode 344. Reverse String

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

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

题意很简单。然后我写了个String result; for(int i=s.length()-1;i--;i>=0){result+=s.charAt(i)}，然后就Time out了。我很郁闷。

然后我就考虑到我可以用两个指针，一个放在字符串头，一个放在字符串尾，两个不停的靠近，将指针所在的字符交换位置。

public class Solution {
    public String reverseString(String s) {
        char[] word = s.toCharArray();
        int i = 0;
        int j = s.length() - 1;
        while (i < j) {
            char temp = word[i];
            word[i] = word[j];
            word[j] = temp;
            i++;
            j--;
        }
        return new String(word);
    }
}

Time Complexity: `O(n)` (Average Case) and `O(n)` (Worst Case) where `n` is the total number character in the input string. The algorithm need to reverse the whole string.
Auxiliary Space: `O(n)` space is used where `n` is the total number character in the input string. Space is needed to transform string to character array.
我看了下有大神用递归的思想是很棒的，为啥我就没有递归的思想呢？？？？？

public class Reverse_String_344 {

	public String reverseString(String s) {
		int length=s.length();
		if(length<=1){
			return s;
		}
		int halfLength=length/2;
		String left=s.substring(0,halfLength);
		String right=s.substring(halfLength,length);
		return reverseString(right)+reverseString(left);
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Reverse_String_344 r=new Reverse_String_344();
		System.out.println(r.reverseString("hello"));
	}

}

Time Complexity: `O(n log(n))` (Average Case) and `O(n * log(n))` (Worst Case) where `n` is the total number character in the input string. The recurrence equation is `T(n) = 2 * T(n/2) + O(n)`. `O(n)` is due to the fact that concatenation function takes linear time. The recurrence equation can be solved to get `O(n * log(n))`.
Auxiliary Space: `O(h)` space is used where `h` is the depth of recursion tree generated which is `log(n)`. Space is needed for activation stack during recursion calls.