本文實例為大家分享了C++合并兩個排序的鏈表,供大家參考,具體內容如下
問題描述
輸入兩個單調遞增的鏈表,輸出兩個鏈表合成后的鏈表,當然我們需要合成后的鏈表滿足單調不減規則。
struct ListNode { int val; struct ListNode *next; ListNode(int x) : val(x), next(NULL) { }};方法一
class Solution {public: ListNode* Merge(ListNode* pHead1, ListNode* pHead2) { ListNode* newList = NULL; //新鏈表頭 ListNode* newListRear = NULL; //新鏈表尾 // 先處理某個鏈表為空的情形 if (pHead1 == NULL){ return pHead2; } if (pHead2 == NULL){ return pHead1; } // 把數值小的結點放入新鏈表,生成頭節點 if (pHead1->val <= pHead2->val){ newList = pHead1; newListRear = pHead1; pHead1 = pHead1->next; }else{ newList = pHead2 ; newListRear = pHead2; pHead2 = pHead2->next; } // 兩表均不空的情形下,遍歷 while (pHead1 != NULL && pHead2 != NULL) { if (pHead1->val <= pHead2->val) { newListRear->next =pHead1; newListRear = pHead1; pHead1 = pHead1->next; }else{ newListRear->next =pHead2; newListRear = pHead2; pHead2 = pHead2->next; } } //某一表為空時,把另一表接入新表表尾 if (pHead1 == NULL) { newListRear->next = pHead2; } if (pHead2 == NULL) { newListRear->next = pHead1; } return newList; }};方法二(遞歸思想)
class Solution {public: ListNode* Merge(ListNode* pHead1, ListNode* pHead2) { if (pHead1 == NULL){ return pHead2; } if (pHead2 == NULL){ return pHead1; } if (pHead1->val <= pHead2->val){ // pHead1為合并后的頭節點 pHead1->next = Merge(pHead1->next, pHead2); return pHead1; }else{ // pHead2 為合并后的頭節點 pHead2->next = Merge(pHead1, pHead2->next); return pHead2; } }};以上就是本文的全部內容,希望對大家的學習有所幫助,也希望大家多多支持武林網。
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