合并單鏈表，輸出單鏈表中間元素，判斷是否有環等

本博文內容為單鏈表相關的筆試題，轉載請註明轉載處，否則必究

1. 合并兩個有序的單鏈表成一個有序的單鏈表

方法分為遞迴實現與非遞迴實現，兩種方法都不額外開闢 記憶體空間

鏈表的資料結構在本部落格的單鏈表逆轉，約瑟夫環等 

遞迴實現：

//遞迴實現合并兩個有序單鏈表LinkNode* merge_list(LinkNode *pHead1, LinkNode *pHead2){if(pHead1==NULL)return pHead2;if(pHead2==NULL)return pHead1;if(pHead1==NULL && pHead2==NULL)return NULL;LinkNode *pMergedHead=NULL;if(pHead1->value<pHead2->value){pMergedHead=pHead1;pMergedHead->next = merge_list(pHead1->next, pHead2);}else {pMergedHead=pHead2;pMergedHead->next=merge_list(pHead1, pHead2->next);}return pMergedHead;}

非遞迴實現：

//非遞迴實現合并兩個有序單鏈表(不額外開闢空間)LinkNode* non_merge_list(LinkNode *pHead1, LinkNode *pHead2){if(pHead1==NULL)return pHead2;if(pHead2==NULL)return pHead1;if(pHead1==NULL && pHead2==NULL)return NULL;LinkNode *pMergedHead = NULL;LinkNode *q=NULL;if(pHead1->value<pHead2->value){pMergedHead=pHead1;pHead1=pHead1->next;}else{pMergedHead=pHead2;pHead2=pHead2->next;}q=pMergedHead;while(pHead1 && pHead2){if(pHead1->value<pHead2->value){q->next=pHead1;pHead1=pHead1->next;}else{q->next=pHead2;pHead2=pHead2->next;}q=q->next;}if(pHead1){while(pHead1){q->next=pHead1;q=q->next;pHead1=pHead1->next;}}if(pHead2){while(pHead2){q->next=pHead2;q=q->next;pHead2=pHead2->next;}}return pMergedHead;}

2 輸出單鏈表中的中間元素(若鏈表節點個數為偶數，則輸出中間兩個的任意一個)

思路：利用兩個指標從前端節點開始遍曆，一個走一步，一個走兩步，當一次走兩步的指標走到鏈表末尾時，此時一次走一步的指標就指向鏈表的中間節點

代碼如下：

LinkNode* print_mid_node(LinkNode *pHead){LinkNode *pOne = pHead, *pTwo = pHead;while(1){pOne = pOne->next;pTwo = pTwo->next->next;if(pTwo==NULL || pTwo->next==NULL)return pOne;}}

3 判斷單戀表是否有環

思路與第二題一樣，只是結束條件不一樣，如果當一次走一步的指標等於一次走兩步的指標時，則表示該鏈表有環

代碼如下：

bool is_circle_list(LinkNode *pHead){LinkNode *pOne = pHead, *pTwo = pHead;while(1){pOne = pOne->next;pTwo = pTwo->next->next;if(pOne == pTwo)return true;if(pTwo==NULL || pTwo->next==NULL)return false;}}