leetcode21 Merge Two Sorted Lists 將兩個有序鏈表組合成一個新的有序鏈表

題目要求

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

翻譯過來就是：將兩個有序的鏈表組合成一個新的有序的鏈表

思路一：循環

在當前兩個鏈表的節點都是非空的狀況下比較大小，較小的添入結果鏈表中而且得到較小節點的下一個節點。這樣比較直至至少遍歷完一個鏈表。再將剩下的鏈表添至結果鏈表的末端

public class MergeTwoSortedLists_21 {

    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode start = new ListNode(0);
        ListNode temp = new ListNode(0);
        start.next = temp;
        while(l1!=null && l2!=null){
            if(l1.val <= l2.val){
                temp.next = l1;
                l1 = l1.next;
            }else{
                temp.next = l2;
                l2 = l2.next;
            }
            temp = temp.next;
        }
        if(l1!=null){
            temp.next = l1;
        }
        if(l2!=null){
            temp.next = l2;
        }
        return start.next.next;
    }
    
    public class ListNode{
        int val;
        ListNode next;
        ListNode(int x) { val = x; }
    }
}

思路二：遞歸

每次比較獲得兩個節點中較小的節點做爲結果返回，並繼續對剩下來的鏈表從新得到較小節點。

public class MergeTwoSortedLists_21 {
    public ListNode mergeTwoLists_recursive(ListNode l1, ListNode l2){
        if(l1==null){
            return l2;
        }else if (l2==null){
            return l1;
        }
        ListNode mergeHead;
        if(l1.val <= l2.val){
            mergeHead = l1;
            mergeHead.next = mergeTwoLists_recursive(l1.next, l2);
        }else{
            mergeHead = l2;
            mergeHead.next = mergeTwoLists(l1, l2.next);
        }
        return mergeHead;
    }
    
    public class ListNode{
        int val;
        ListNode next;
        ListNode(int x) { val = x; }
    }
}

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