LeetCode 分隔链表

题目链接：https://leetcode-cn.com/problems/partition-list/

题目大意

　　略。

分析

　　空间复杂度 O(1) 的做法蛮有意思的，另外加头结点可以少写很多代码。

代码如下

/**
 * 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 *leftH, *leftT, *rightH, *rightT, *p = head;
        leftH = leftT = new ListNode(0);
        rightH = rightT = new ListNode(0);
        
        while(p != NULL) {
            if(p->val < x) {
                leftT->next = p;
                leftT =leftT->next;
                
            }
            else if(p->val >= x) {
                rightT->next = p;
                rightT = rightT->next;
            }
            p = p->next;
        }
        
        leftT->next = rightH->next;
        rightT->next = NULL;
        
        return leftH->next;
    }
};