leetcode430. Flatten a Multilevel Doubly Linked Li

gxyz 發(fā)布于2019-08-16 18:29 / 1581人閱讀

摘要：步驟如下代碼如下思路二循環(huán)上面的思路同樣可以通過循環(huán)的方式來解決。基本步驟如下代碼如下思路減少遍歷次數(shù)之前的兩種思路，都會(huì)出現(xiàn)大量的重復(fù)遍歷，重復(fù)遍歷和葉子節(jié)點(diǎn)的深度成正相關(guān)，可以想方法將重復(fù)遍歷的次數(shù)減少。

題目要求

You are given a doubly linked list which in addition to the next and previous pointers, it could have a child pointer, which may or may not point to a separate doubly linked list. These child lists may have one or more children of their own, and so on, to produce a multilevel data structure, as shown in the example below.

Flatten the list so that all the nodes appear in a single-level, doubly linked list. You are given the head of the first level of the list.

 

Example:

Input:
 1---2---3---4---5---6--NULL
         |
         7---8---9---10--NULL
             |
             11--12--NULL

Output:
1-2-3-7-8-11-12-9-10-4-5-6-NULL

思路一：遞歸實(shí)現(xiàn)深度優(yōu)先遍歷

從深度優(yōu)先遍歷的角度來看，每次遇到一個(gè)包含子節(jié)點(diǎn)中間雙鏈表節(jié)點(diǎn)，就遞歸的調(diào)用展開方法將其展開，并將展開的結(jié)果插入到當(dāng)前節(jié)點(diǎn)的后面。這里需要注意雙鏈表前節(jié)點(diǎn)前后指針的變更。步驟如下：

Step1:
 1---2---3---4---5---6--NULL
         |
         7---8---9---10--NULL
             |
             11--12--NULL

Step2:
1---2---3---4---5---6--NULL
         |
         7---8---11--12--9---10--NULL
         
Step3:
1---2---3---7---8---11--12--9---10--4---5---6--NULL

代碼如下：

    public Node flatten(Node head) {
        if(head == null) return head;
        Node tmp = head;
        while(tmp != null) {
            if(tmp.child != null) {
                Node child = flatten(tmp.child);
                tmp.child = null;
                Node next = tmp.next;
                tmp.next = child;
                child.prev = tmp;
                while(child.next != null) {
                    child = child.next;
                }
                child.next = next;
                if(next != null) {
                    next.prev = child;
                }
                tmp = next;
            }else {
                tmp = tmp.next;

            }
        }
        return head;
    }

思路二：循環(huán)

上面的思路同樣可以通過循環(huán)的方式來解決。每遇到一個(gè)有子節(jié)點(diǎn)的雙鏈表節(jié)點(diǎn)，就將其子節(jié)點(diǎn)的頭和尾拼接到父節(jié)點(diǎn)的雙鏈表上，使其看上去是一個(gè)新的雙鏈表。再對(duì)雙鏈表的下一個(gè)節(jié)點(diǎn)進(jìn)行判斷。基本步驟如下：

Step1:
 1---2---3---4---5---6--NULL
         |
         7---8---9---10--NULL
             |
             11--12--NULL

Step2:
1---2---3---7---8---9---10---4---5---6--NULL
             |
             11--12--NULL
         
Step3:
1---2---3---7---8---11--12--9---10--4---5---6--NULL

代碼如下：

    public Node flatten(Node head) {
        if(head == null) return null;
        
        Node tmp = head;
        while(tmp != null) {
            if(tmp.child != null) {
                
                Node child = tmp.child;
                tmp.child = null;
                
                Node next = tmp.next;
                tmp.next = child;
                child.prev = tmp;
                while(child.next != null) {
                    child =  child.next;
                }
                
                if(next != null) {
                    child.next = next;
                    next.prev = child;
                }
            }
            tmp = tmp.next;
        }
        return head;
    }

思路3：減少遍歷次數(shù)

之前的兩種思路，都會(huì)出現(xiàn)大量的重復(fù)遍歷，重復(fù)遍歷和葉子節(jié)點(diǎn)的深度成正相關(guān)，可以想方法將重復(fù)遍歷的次數(shù)減少。其實(shí)，我們可以看見，無論我們何時(shí)將子節(jié)點(diǎn)展開，并拼接回父節(jié)點(diǎn)的雙鏈表中，子節(jié)點(diǎn)展開的雙鏈表的頭結(jié)點(diǎn)是固定的，并且可以用父節(jié)點(diǎn)訪問到。而尾節(jié)點(diǎn)必須通過重復(fù)遍歷來查找并拼接。因此，如果每次都將展開后的尾節(jié)點(diǎn)返回，就可以無需重復(fù)遍歷將展開的子節(jié)點(diǎn)拼接回父節(jié)點(diǎn)。代碼如下：

    public Node flatten(Node head) {
        flattenAndReturnTail(head);
        return head;
    }
    
    public Node flattenAndReturnTail(Node head) {
        if(head == null) return null;
        if(head.child == null) {
            if(head.next == null) return head;
            return flattenAndReturnTail(head.next);
        }else {
            Node child = head.child;
            head.child = null;
            
            Node next = head.next;
            Node childTail = flatten(child);
            head.next  = child;
            child.prev = head;
            if(next != null) {
                childTail.next = next;
                next.prev = childTail;
                return flattenAndReturnTail(next);
            }
            return childTail;
        }
    }