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.
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class Solution {
public:
ListNode* partition(ListNode* head, int x) {
ListNode *small=NULL, *greater=NULL, *smallCur=NULL, *greaterCur=NULL;
if(head==NULL) return head;
for(ListNode *cur=head; cur!=NULL;) {
ListNode *next = cur->next;
if(cur->val < x) {
if(small==NULL) {
small=cur;
} else {
smallCur->next=cur;
}
smallCur=cur;
smallCur->next = NULL;
} else {
if(greater==NULL) {
greater=cur;
} else {
greaterCur->next=cur;
}
greaterCur=cur;
greaterCur->next=NULL;
}
cur = next;
}
if(small!=NULL) {
smallCur->next=greater;
} else {
small=greater;
}
return small;
}
};
本文介绍了一种链表分区算法,该算法将链表中小于给定值x的所有节点置于大于等于x的节点之前,同时保持各分区内的相对顺序不变。例如,对于链表1->4->3->2->5->2 和 x = 3,算法返回 1->2->2->4->3->5。通过遍历链表,并创建两个新的链表分别存储小于x和大于等于x的节点,最终将两链表合并。

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