题目
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.
解题之法
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *partition(ListNode *head, int x) {
if (!head) return head;
ListNode *dummy = new ListNode(-1);
ListNode *newDummy = new ListNode(-1);
dummy->next = head;
ListNode *cur = dummy, *p = newDummy;
while (cur->next) {
if (cur->next->val < x) {
p->next = cur->next;
p = p->next;
cur->next = cur->next->next;
p->next = NULL;
} else {
cur = cur->next;
}
}
p->next = dummy->next;
return newDummy->next;
}
};
分析
这道题要求我们划分链表，把所有小于给定值的节点都移到前面，大于该值的节点顺序不变，相当于一个局部排序的问题。
可以将所有小于给定值的节点取出组成一个新的链表，此时原链表中剩余的节点的值都大于或等于给定值，只要将原链表直接接在新链表后即可。
