LeetCode刷题笔记（链表）：partition-list

分手后的思念是犯贱 2022-06-01 01:40 167阅读 0赞

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*  **转载请注明作者和出处：[http://blog.csdn.net/u011475210][http_blog.csdn.net_u011475210]**
 *  **代码地址：[https://github.com/WordZzzz/Note/tree/master/LeetCode][https_github.com_WordZzzz_Note_tree_master_LeetCode]**
 *  **刷题平台：[https://www.nowcoder.com/ta/leetcode][https_www.nowcoder.com_ta_leetcode]**
 *  **题  库：Leetcode经典编程题**
 *  **编  者：WordZzzz**

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*   *  题目描述
     *  解题思路
     *  C版代码实现

## 题目描述 ##

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,  
Given1->4->3->2->5->2and x = 3,  
return1->2->2->4->3->5.

## 解题思路 ##

牛客网上面的答案都是新建两个链表，小于x的放到一个链表里面，不小于的放到另一个链表里面，这种答案感觉好没劲哦。所以最后我采用的是O(n)时间复杂度，O(1)空间复杂度的解法。

具体说来，还是用快慢指针遍历链表，slow指向连续小于x的最后一个元素，fast指向当前元素不小于x但是下个元素小于x的元素。理解清楚这两个指针的对应关系之后，我们很容易将fast指向的元素的下一个元素追加到slow之后，同时更新slow和fast的值。

## C++版代码实现 ##

/**
     * 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) {
            if(head == NULL)
                return NULL;
    
            ListNode *dummy = new ListNode(0);
            dummy->next = head;
            ListNode *fast = dummy;
            ListNode *slow = dummy;
            while(fast != NULL && fast->next != NULL){
                if(fast->next->val >= x)
                    fast = fast->next;
                else{
                    if(fast == slow){
                        fast = fast->next;
                        slow = slow->next;
                    }
                    else{
                        ListNode *tmp = fast->next;
                        fast->next = tmp->next;
                        tmp->next = slow->next;
                        slow->next = tmp;
                        slow = slow->next;
    
                    }
                }
            }
            return dummy->next;
        }
    };

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[http_blog.csdn.net_u011475210]: http://blog.csdn.net/u011475210
[https_github.com_WordZzzz_Note_tree_master_LeetCode]: https://github.com/WordZzzz/Note/tree/master/LeetCode
[https_www.nowcoder.com_ta_leetcode]: https://www.nowcoder.com/ta/leetcode