链表题目---3 合并两个有序单链表和分割链表

合并两个有序单链表

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/*** Definition for singly-linked list.* struct ListNode {*     int val;*     struct ListNode *next;* };*/struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2){if(l1 == NULL){return l2;}else if(l2 == NULL){return l1;}struct ListNode *cur1 = l1;struct ListNode *cur2 = l2;struct ListNode *result =NULL;  //合并后的新链表struct ListNode *result_tail =NULL; //result 链表的最后一个结点while(cur1!=NULL&&cur2!=NULL){if(cur1->val <= cur2->val ){//把cur1的结点尾插到resultif(result_tail !=NULL){  //往最后一个结点插result_tail->next=cur1;result_tail=cur1;}else{result =cur1;result_tail=cur1;}cur1=cur1->next;		//往后移}else{//把cur2的结点尾插到resultif(result_tail !=NULL){result_tail->next=cur2;result_tail=cur2;}else{result =cur2;result_tail=cur2;}cur2=cur2->next;       //往后移}}//一个链表已经处理完了if(cur1 != NULL){result_tail->next=cur1;}else{result_tail->next=cur2;}return result;
}

分割链表

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/*
struct ListNode {int val;struct ListNode *next;ListNode(int x) : val(x), next(NULL) {}
};*/
class Partition {
public:ListNode* partition(ListNode* pHead, int x) {ListNode *lt =NULL;     //  <x的第一个结点ListNode *lt_tail =NULL;    //<x 的最后一个结点ListNode *ge =NULL; // >=x的第一个结点ListNode *ge_tail = NULL; // >=x 的最后一个结点ListNode *cur =pHead;while(cur != NULL){ListNode *next =cur->next;cur->next = NULL;if(cur->val< x){if(lt_tail == NULL){lt = lt_tail =cur;}else{lt_tail->next = cur;lt_tail = cur;}}else{if(ge_tail == NULL){ge = ge_tail =cur;}else{ge_tail->next = cur;ge_tail = cur;}}cur = next;}if(lt_tail == NULL){return ge;}lt_tail->next = ge;return lt;}
};