Merge Two Sorted Lists

LintCode LeetCode LinkedList

这道题暴力法的思路很straightforwad :
- 先找到新merged list 的head，并将head.next = null
- 继续在两个链表的头中找最小的，一个一个地接到新的list后面

    /**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null){
            return l2;
        }
        if (l2 == null){
            return l1;
        }
        ListNode dummy = new ListNode(0);
        //find the head of the merged list
        if (l1.val <= l2.val){
            dummy.next = l1;
            l1 = l1.next;
        } else {
            dummy.next = l2;
            l2 = l2.next;
        }
        ListNode mergedHead = dummy.next;
        mergedHead.next = null;
        // 这一部分这么写就不行：
//         ListNode dummy = new ListNode(0);
//         ListNode mergedHead = dummy.next;
//         if (l1.val <= l2.val){
//             mergedHead = l1;
//             l1 = l1.next;
//         } else {
//             mergedHead = l2;
//             l2 = l2.next;
//         }
//         mergedHead.next = null;
        //find the next item of the merged list one by one
        while (l1 != null && l2 != null){
            if (l1.val <= l2.val){
                mergedHead.next = l1;
                l1 = l1.next;
            } else {
                mergedHead.next = l2;
                l2 = l2.next;
            }   
            mergedHead = mergedHead.next;
            mergedHead.next = null;      
        }
        //if one list is null, we can append the rest part of the other list since it is sorted.
        if (l1 == null){
            mergedHead.next = l2;
        } else if (l2 == null){
            mergedHead.next = l1;
        }
        return dummy.next;
    }
}

还有一种思路是先不确定新链表的头是什么，随便从两个参数链表中拿出一个比如l1接到dummy node后面，然后再比较现在l1, l2的头的大小. 如果l2比l1小，就在l1前面加入l2的当前元素；如果l1比l2小，就只需要移动l1和prev.

public class Solution {
    /**
     * @param ListNode l1 is the head of the linked list
     * @param ListNode l2 is the head of the linked list
     * @return: ListNode head of linked list
     */
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        // write your code here
        //dummy is still, prev is moving
        ListNode dummy = new ListNode(0);
        ListNode prev = dummy;
        //we casually choose one list to start
        dummy.next = l1;
        while (l1 != null && l2 != null){
            if (l2.val < l1.val){
                ListNode temp = l2.next;
                l2.next = prev.next;
                prev.next = l2;
                l2 = temp;
                prev = prev.next;
            } else {
                l1 = l1.next;
                prev = prev.next;
            }
        }
        if (l1 != null){
            prev.next = l1;
        } else if (l2 != null){
            prev.next = l2;
        }
        return dummy.next;
    }
}