链表－leetcode 86 Partition List

原题链接：Partition List

题解：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* partition(ListNode* head, int x) {        /*        思路：类似与快排的partition过程，需要有两个指针分别指向小值和大值的末尾，小于就换位置，大于就下一个        Time Complexity:O(N)            Space Complexity:O(1)        */        if(!head || !head->next)return head;        ListNode* res=new ListNode(0);        res->next=head;        ListNode* small=res,*max=NULL;        bool flag=false;        while(head){            if(head->val<x){                if(small->next==head){                    small=head;                    head=head->next;                }                else{                    max->next=head->next;                    head->next=small->next;                    small->next=head;                    small=head;                    head=max->next;                }            }            else{                if(!flag){                    max=head;                    flag=true;                    head=head->next;                }                else{                    max=head;                    head=head->next;                }            }        }        return res->next;    }};