【牛客網】合併兩個排序的鏈表

• 題目：
  輸入兩個單調遞增的鏈表，輸出兩個鏈表合成後的鏈表，當然我們需要合成後的鏈表滿足單調不減規則。

• 超時的代碼：

class Solution {
public:
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
    {
     ListNode* k;
    if (pHead1 == NULL && pHead2 == NULL)return NULL;
    if (pHead1 == NULL)return pHead2;
    if (pHead2 == NULL)return pHead1;
    ((pHead1->val) > (pHead2->val)) ? k = pHead1 : k = pHead2;
    ListNode* m=k;
    while (pHead1 != NULL && pHead2 != NULL){
        if ((pHead1->val) >= (pHead2->val)){
            k->next= pHead2;
            pHead2 = pHead2->next;
            k = k->next;
        }
        else{
            k->next = pHead1;
            pHead1 = pHead1->next;
            k = k->next;
        }
    }
    if (pHead1 ==NULL && pHead2 == NULL){
        k->next = NULL;
    }
    else if (pHead1 == NULL){
        while (pHead2 != NULL){
            k->next = pHead2;
            pHead2 = pHead2->next;
            k = k->next;
        }
        k->next = NULL;
    }
    else{
        while (pHead1 != NULL){
            k->next = pHead1;
            pHead1 = pHead1->next;
            k = k->next;
        }
        k->next = NULL;
    }
    return m;
    }
};//這裏沒有考慮到用遞歸的思想

• 劍指Offer上的代碼

class Solution {
public:
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
    {
        if (pHead1 == NULL)
           return pHead2;
        else if (pHead2 == NULL)
           return pHead1;
        ListNode* pMergedHead = NULL;
        if (pHead1->val < pHead2->val){
             pMergedHead = pHead1;
             pMergedHead->next = Merge(pHead1->next, pHead2);
        }
        else{
            pMergedHead = pHead2;
            pMergedHead->next = Merge(pHead1, pHead2->next);
        }
        return pMergedHead;
    }
};//這裏用遞歸算法進行歸併，太好了