2. 程式人生 > >21. Merge Two Sorted Lists 一道基礎題的兩種精煉解法

21. Merge Two Sorted Lists 一道基礎題的兩種精煉解法

阿新 • • 發佈：2019-01-19

方法一：遞迴

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        if(l1 == NULL) return l2;
        if(l2 == NULL) return l1;
        
        if(l1->val > l2->val){
            l2->next = mergeTwoLists(l1, l2->next);
            return l2;
        }else{
            l1->next = mergeTwoLists(l1->next, l2);
            return l1;
        }
    }
};

法二：遞迴的精煉程式碼： 
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode dummy(1);
        ListNode* tail = &dummy;// 妙， 設定指向尾的指標， 並且對dummy取地址，不用new了，函式結束後自動釋放
        while(l1 && l2){
            if(l1->val > l2->val){
                tail->next = l2;
                l2 = l2->next;
            }else{
                tail->next = l1;
                l1 = l1->next;
            }
            tail = tail->next;
        }
        //tail->next = l1 ? l1 : l2;
        l1 == NULL? tail->next = l2: tail->next = l1;
        return dummy.next;
    }
};


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