ReverseString.java

/**
 * <p>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 <code>s</code> 的形式给出。</p>
 *
 * <p>不要给另外的数组分配额外的空间，你必须<strong><a href="https://baike.baidu.com/item/原地算法" target="_blank">原地</a>修改输入数组</strong>、使用 O(1) 的额外空间解决这一问题。</p>
 *
 * <p>&nbsp;</p>
 *
 * <p><strong>示例 1：</strong></p>
 *
 * <pre>
 * <strong>输入：</strong>s = ["h","e","l","l","o"]
 * <strong>输出：</strong>["o","l","l","e","h"]
 * </pre>
 *
 * <p><strong>示例 2：</strong></p>
 *
 * <pre>
 * <strong>输入：</strong>s = ["H","a","n","n","a","h"]
 * <strong>输出：</strong>["h","a","n","n","a","H"]</pre>
 *
 * <p>&nbsp;</p>
 *
 * <p><strong>提示：</strong></p>
 *
 * <ul>
 * <li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
 * <li><code>s[i]</code> 都是 <a href="https://baike.baidu.com/item/ASCII" target="_blank">ASCII</a> 码表中的可打印字符</li>
 * </ul>
 * <div><div>Related Topics</div><div><li>递归</li><li>双指针</li><li>字符串</li></div></div><br><div><li>👍 518</li><li>👎 0</li></div>
 */

package leetcode9;

public class ReverseString {

    public static void main(String[] args) {
        Solution solution = new ReverseString().new Solution();
    }

    class Solution {
        public void reverseString(char[] s) {
            for (int left = 0, right = s.length - 1; left < right; left++, right--) {
                char temp = s[left];
                s[left] = s[right];
                s[right] = temp;
            }
        }
    }
}