[LeetCode] Partition List 分割单链表

来源：互联网发布：保定网络托管编辑：程序博客网时间：2024/05/22 06:47

给你一个单链表和一个整数x，调整单链表的元素的位置，使得前半部分的值小于x，后半部分的值大于x。

下面的算法的思路和单链表快速排序的思路是一样的，该算法有可能改变节点的相对顺序。

ListNode *partition(ListNode *head, int x) {        ListNode* slow = head;        ListNode* fast = head;        while(fast)        {                if(fast->val < x) // note the less-than-equal operator here                // Input: 5->4->3 and x = 4                // Output: 4->3->5                {                        swap(slow->val, fast->val);                        slow = slow->next;                }                fast = fast->next;        }         return head;}

2. 下面的算法可以保持相对顺序不变。

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x. You should preserve the original relative order of the nodes in each of the two partitions. For example, Given 1->4->3->2->5->2 and x = 3, return 1->2->2->4->3->5.

思路很简单：先找出链表的长度和链表的尾节点，然后遍历链表，如果当前节点的值小于target，跳到下一个节点，如果当前节点的值大于等于target，则把这个节点移动到链表的末端。

/** * 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) {        ListNode *p= new ListNode(0);        p->next = head;        head = p; // used to save the result head.        ListNode *last=head; // used to get the last node                 if (head->next==NULL){return head->next;}                 int n=0; //length of the list        while (last->next!=NULL){ last=last->next; n++; } //get the length and last node                 while (n>0){  // in case  of non-stop loop, count n.            if (p->next->val < x){  // val<x keep the node                p=p->next;                n--;            }else{                  // val>=x move to last                last->next = new ListNode(p->next->val);    // add node to the last                last = last->next;                p->next = p->next->next;                    //delete current node                n--;            }        }        return head->next;  //the 1st node is elmininated     }};

0 0