LeetCode 刷题【86. 分隔链表】

86. 分隔链表

自己做

解：双指针法

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode() : val(0), next(nullptr) {}*     ListNode(int x) : val(x), next(nullptr) {}*     ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* partition(ListNode* head, int x) {ListNode* p = head;ListNode* q = head;ListNode* r1 = head;ListNode* r2 = head;while(q != nullptr && q->val < x){          //分割线p = q;                  //分割线前一位q = q->next;r1 = q;                 //移动结点前一位if(q != nullptr)r2 = q->next;           //移动结点elser2 = q;                 //如果q为空说明这个链表的边界就是末尾}while(r2 != nullptr){if(r2->val < x){   //找到要移动的目标值if(q == head){      //在头结点前插入，涉及头结点的更改head = r2;r1->next = r2->next;r2->next = q;p = head;r2 = r1->next;}else{p->next = r2;r1->next = r2->next;r2->next = q;p = p->next;r2 = r1->next;}}else{                   //继续找r1 = r2;r2 = r2->next;}}return head;}};