Leetcode 86. Partition List （链表好题）

86. Partition List
    Medium

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

解法1：
注意不能直接用node=node->next。

        ListNode *tmp = node->next;
        //node = node->next;
        node->next = NULL;
        node = tmp;
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以 [1,4,3,2,5,2] 为例，5后面是2，如果不断开，就会形成环。

/**
 * 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 *dummy1 = new ListNode(0), *dummy2 = new ListNode(0);
        ListNode *node1 = dummy1, *node2 = dummy2, *node = head;
        while (node) {
            if (node->val < x) {
                node1->next = node;
                node1 = node1->next;
            } else {
                node2->next = node;
                node2 = node2->next;
            }
            ListNode *tmp = node->next;
            //node = node->next;
            node->next = NULL;
            node = tmp;
        }
        node1->next = dummy2->next;
        ListNode *res = dummy1->next;
        delete(dummy1); delete(dummy2);
        return res;
    }
};
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