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21. Merge Two Sorted Lists

21. Merge Two Sorted Lists

作者: 番茄晓蛋 | 来源:发表于2017-06-10 12:06 被阅读8次

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    Total Accepted: 221656
    Total Submissions: 571422
    Difficulty: Easy
    Contributor: LeetCode

    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.

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    ** 解题思路**
    recursive 递归解法。 类似merge two sorted array

    /**
     * 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) {
          // use recursive 
          if (l2 == null)  return l1;
          if (l1 == null)  return l2;
          
          ListNode mergeHead;
          
          if (l1.val < l2.val) {
              mergeHead = l1;
              mergeHead.next = mergeTwoLists(l1.next, l2);
          } else {
              mergeHead = l2;
              mergeHead.next = mergeTwoLists(l1, l2.next);
          }
          return mergeHead;
        }
        
        public ListNode mergeTwoListsFailed(ListNode l1, ListNode l2) {
            ListNode p1 = l1;
            ListNode p2 = l2 ;
            ListNode p = new ListNode(0);
            
            
            while (p1 != null && p2 != null) {
                // p.next = (p1.val > p2.val) ? new ListNode(p1.val) : new ListNode(p2.val);
                if (p1.val > p2.val) {
                    p.next = new ListNode(p1.val);
                    p1 = p1.next;
                } else {
                    p.next = new ListNode(p2.val);
                    p2 = p2.next;
                }
             
            }
            
            while (p2 != null) {
              p.next = new ListNode(p2.val);
              p2 = p2.next;
             
            }
            
            while (p1 != null) {
                p.next = new ListNode(p1.val);
                p1 = p1.next;
              
            }
            return p;
        }
    }
    

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