題目:
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.
分析:
在一個鏈表中,將值小於x的node移到前面去。注意,不能改變node之間原本的相對位置。比如x=3, 1原本出現在2的前面,那麼修改後的鏈表中,值爲1的node也必須出現在值爲2的前面。
思路1:
遍歷,將val小於x的node移到前面去,通過改變指針的指向來實現。這樣做有個問題,就是,指針的指向改變比較麻煩,同時,需要提前記錄很多值的位置。
代碼1:
/**
* 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* place = new ListNode(0);
ListNode* h = place;
ListNode* place_next;
ListNode* now = head;
ListNode* now_next;
ListNode* pre = place;
place->next = head;
while(now)
{
now_next = now->next;
if(now->val < x )
{
place_next = place->next;
place->next = now;
if(place_next != now)
{
now->next = place_next;
pre->next = now_next;
}
place = now;
}
else
{
pre = now;
}
now = now_next;
}
return h->next;
}
};
【參考思路 引自:https://leetcode.com/discuss/21032/very-concise-one-pass-solution】
將值小於x的node弄出來,作爲一條鏈表,將另外一部分作爲另外一條鏈表。最後將兩個鏈表連起來就可以了。相當簡單啊,沒有複雜的指針變化問題。喜歡!!
代碼:
ListNode *partition(ListNode *head, int x) {
ListNode node1(0), node2(0);
ListNode *p1 = &node1, *p2 = &node2;
while (head) {
if (head->val < x)
p1 = p1->next = head;
else
p2 = p2->next = head;
head = head->next;
}
p2->next = NULL;
p1->next = node2.next;
return node1.next;
}