- 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;
以 [1,4,3,2,5,2] 为例,5后面是2,如果不断开,就会形成环。
cpp
/**
* 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;
}
};