递归算法

Posted 2022-08-23 kaleidoscope

为了确保递归函数不会导致无限循环，它应具有以下属性：

• 一个简单的基本案例（basic case）（或一些案例） —— 能够不使用递归来产生答案的终止方案。
• 一组规则，也称作递推关系（recurrence relation），可将所有其他情况拆分到基本案例。

1、以相反的顺序打印字符串

/**
   输入：["h","e","l","l","o"]
   输出：["o","l","l","e","h"]
 */
public class PrintReverseDemo 

    public static void main(String[] args) 
        printReverse("apple".toCharArray());
    

    private static void printReverse(char[] str) 
        helper(0, str);
    

    private static void helper(int index, char[] str) 

        if (null == str || index >= str.length) 
            return;
        
        helper(index + 1, str);
        System.out.print(str[index]);
    

 

2、两两交换链表中的节点

/**
 * 给定 1->2->3->4, 你应该返回 2->1->4->3.
 */
public class SwapNodeDemo 

    public static void main(String[] args)

        ListNode head = makeListNode();

        System.out.println(head);
        System.out.println(swapPairs2(head));



    /**
     * 递归
     * @param head
     * @return
     */
    public static ListNode swapPairs(ListNode head) 
        if (head == null || head.next == null) 
            return head;
        
        
        ListNode p = head.next;
        head.next = swapPairs(head.next.next);
        p.next = head;
        return p;
    

    /**
     * 非递归
     * @param head
     * @return
     */
    public static ListNode swapPairs2(ListNode head) 
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode curr = dummy;
        while (curr.next != null && curr.next.next != null) 
            ListNode first = curr.next;
            ListNode second = curr.next.next;

            // swap two nodes
            first.next = second.next;
            second.next = first;
            curr.next = second;

            // update to next iteration
            curr = curr.next.next;
        
        return dummy.next;
    

    public static ListNode makeListNode() 
        ListNode one = new ListNode(1);
        ListNode two = new ListNode(2);
        ListNode three = new ListNode(3);
        ListNode four = new ListNode(4);

        one.next = two;
        two.next = three;
        three.next = four;

        return one;