英文原文
Write code to partition a linked list around a value x, such that all nodes less than x come before all nodes greater than or equal to x. If x is contained within the list, the values of x only need to be after the elements less than x (see below). The partition element x can appear anywhere in the "right partition"; it does not need to appear between the left and right partitions.
Example:
Input: head = 3->5->8->5->10->2->1, x = 5 Output: 3->1->2->10->5->5->8
中文题目
给你一个链表的头节点 head 和一个特定值 x ，请你对链表进行分隔，使得所有 小于 x 的节点都出现在 大于或等于 x 的节点之前。
你不需要 保留 每个分区中各节点的初始相对位置。
示例 1：
输入：head = [1,4,3,2,5,2], x = 3 输出：[1,2,2,4,3,5]
示例 2：
输入：head = [2,1], x = 2 输出：[1,2]
提示：
- 链表中节点的数目在范围
[0, 200]内 -100 <= Node.val <= 100-200 <= x <= 200
通过代码
高赞题解
if (!head || !head->next) return head;
struct ListNode* h1 = (struct ListNode*)malloc(sizeof(struct ListNode)), *t1 = h1;
struct ListNode* h2 = (struct ListNode*)malloc(sizeof(struct ListNode)), *t2 = h2;
t1->next = NULL; t2->next = NULL;
while (head){
if (head->val < x){
t1->next = head;
t1 = t1->next;
}
else{
t2->next = head;
t2 = t2->next;
}
head = head->next;
}
t1->next = h2->next;
t2->next = NULL;
return h1->next;统计信息
| 通过次数 | 提交次数 | AC比率 |
|---|---|---|
| 26835 | 41232 | 65.1% |
提交历史
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