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.
Example:
Input: head = 1->4->3->2->5->2, x = 3
Output: 1->2->2->4->3->5
AC代码
class Solution {
public:
ListNode* partition(ListNode* head, int x) {
if (!head || !(head->next)) return head;
ListNode *pre = new ListNode(INT_MIN), *ret = pre, *cur = pre;
pre->next = head;
while (cur->next) {
if (cur->next->val >= x)
cur = cur->next;
else {
if (cur == pre) {
pre = pre->next;
cur = cur->next;
}
else {
ListNode* tmp = cur->next;
cur->next = tmp->next;
tmp->next = pre->next;
pre->next = tmp;
pre = pre->next;
}
}
}
return ret->next;
}
};
总结
循规蹈矩的链表基本操作题
