86. Partition List ------------M

Given the head of 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.

 

Example 1:

Input: head = [1,4,3,2,5,2], x = 3
Output: [1,2,2,4,3,5]

Example 2:

Input: head = [2,1], x = 2
Output: [1,2]

 

Constraints:

• The number of nodes in the list is in the range [0, 200].
• -100 <= Node.val <= 100
• -200 <= x <= 200

A:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* partition(ListNode* head, int x) {
        ListNode hSmall, hBig;
        auto tailS = &hSmall, tailB = &hBig;
        while (head) {
            auto tmp = head;
            head = head->next;
            tmp->next = nullptr;
            if (tmp->val < x) {
                tailS->next = tmp;
                tailS = tailS->next;
            } else {
                tailB->next = tmp;
                tailB = tailB->next;
            }
        }
        tailS->next = hBig.next;
        return hSmall.next;
    }
};

Mistakes:
1: detach 一个node， 并将之放到新的list的tail上时，， 忘记把它的next节点设为null

2: 由于命名的原因（更主要的是，开始没有思考清楚，导致） 最后把tail.next = header 了。


 --------第二遍， ---思路同上----------
 Mistakes:
1：    lowerTail.next = higherHeader.next;  刚开始，SB了，写成lowerHeader.next = higherHeader.next;

Error：
 while(head!= null){
            ListNode tmp = head;
            tmp.next = null;
            head = head.next;
          
这里，这样写，顺序是不对的。 应该先 head = head.next之后，再把tmp的指针置null