给定一个单链表和数值x,划分链表使得所有小于x的节点排在大于等于x的节点之前。
你应该保留两部分内链表节点原有的相对顺序。
2.完全没有思路..........
3./**
* Definition of ListNode
* class ListNode {
* public:
* int val;
* ListNode *next;
* ListNode(int val) {
* this->val = val;
* this->next = NULL;
* }
* }
*/
class Solution {
public:
ListNode *partition(ListNode *head, int x) {
ListNode * small = new ListNode(0); //小于x
ListNode * large = new ListNode(0); //大于x
ListNode * lastSmall = small,* lastLarge = large;
ListNode * cur = head;
while(cur){
if(cur->val < x){
lastSmall->next = cur;
lastSmall = lastSmall->next;
}else{
lastLarge->next = cur;
lastLarge = lastLarge->next;
}
cur = cur->next;
lastSmall->next = lastLarge->next = NULL;
}
lastSmall->next = large->next;
return small->next;
}
};
4.感想
完全不会啊这道题·.有种百度看了答案都不懂得感觉..问问同学.也感觉模模糊糊....
只能先放下这道题.等在学习后再回来看看.