leetcode:Partition List

时间:2023-01-31 15:55:41

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.

分析:

--->题意:给定一个单链表和一个x,把链表中小于x的放到前面,大于等于x的放到后面,每部分元素的原始相对位置不变。

--->思路:遍历一遍链表,把小于x的都挂到part1后,把大于等于x的都放到part2后,最后再把大于等于的链表挂到小于链表的后面就可以了。

代码:

/**
* 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) {
ListNode *part1 = new ListNode(0), *part2 = new ListNode(0), *node1, *node2;
node1 = part1;
node2 = part2;
while (head) {
if (head->val < x) {
node1->next = head;
node1 = node1->next;
} else {
node2->next = head;
node2 = node2->next;
}
head = head->next;
}
node2->next = NULL;
node1->next = part2->next;
return part1->next;
}
};