Partition List

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：1) 先过一遍，找到Tail节点，然后遇到大的就往后放 缺点是：需要2遍的遍历

        2) 维持两个头节点，分别是less_head和greater_head,指向小于x和大于x的链表，原来的List内部的Node只是指针在发生变化。技巧：dummy node的技巧很重要，因为不需要判断head是否为空的问题

教训：1）greater_head指向的链表最后p->next = NULL，否则会导致死循环，这也是刚开始写的时候老是TLE的原因  2）不要用很长的变量名，应该多用ans dp数组 h1 h2 p1 p2这样的变量名，虽然牺牲了一点可读性，但是代码会很简洁，写变量名本身就不会出错了

class Solution {
public:
         ListNode *partition(ListNode *head, int x) {
             if (!head)
                 return NULL;
             ListNode * less_head = NULL, *less_p = NULL;
             ListNode * greater_head = NULL,*greater_p = NULL;
 
             for(ListNode * p = head; p; p = p->next){
                 if (p->val < x){
                    if (!less_p){
                        less_head = p;
                        less_p = p;
                     }else{
                        less_p->next = p;
                        less_p = less_p->next;
                     }
                 }else{ 
                     if (!greater_p){ 
                         greater_head = p;
                         greater_p = p;
                     }else{
                         greater_p->next = p;
                         greater_p = greater_p->next;
                    }
                 }
             }
            if (less_head){
                 less_p->next = greater_head;
                  if (greater_p){
                    greater_p->next = NULL;
                  }
                 return less_head;
             }
 
             return greater_head;
         }
};

 

  上边的代码有很多冗余的地方，不够干净漂亮 1）head是否需要判断  2） 赋值怎么缩减  3）dummy node技巧的使用

class Solution {
public:
         ListNode *partition(ListNode *head, int x) {
             if (!head)
                 return NULL;
             //注意这里构造函数next为空，那么应该明确的构造
             ListNode h1(0), *p1 = &h1;
             ListNode h2(0),*p2 = &h2;
 
             for(ListNode * p = head; p; p = p->next){
                 if (p->val < x){
                     p1 = p1->next = p;
                 }else{
                    p2 = p2->next = p;
                 }
             }
             if (h1.next){
                 p1->next = h2.next;
                 p2->next = NULL;
                 return h1.next;
             }
             return h2.next;
         }
};