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
            else
                r2 = 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;
    
    }

};