LeetCode
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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
.
题意:将值小于x的结点放到前边,值大于等于x的结点放到后边,两部分都保持相对顺序不变。
本来的做法是在原链表上改动,调整先后顺序,然额改来改去自己就很懵。。。越来越懵。。。拜读了评论区的做法,我果然是个渣渣嘤嘤嘤QAQ
解法:新建两个链表,将小于x值的结点放入链表1,大于等于x的放入链表2,然后将1的next指向2,将2尾部的next指向NULL,最后返回1。时间复杂度O(n),空间复杂度O(n)
/** * 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 node1(0), node2(0); ListNode *p1 = &node1, *p2 = &node2; while (head) { if (head->val < x) { p1->next = head; p1 = p1->next; } else { p2->next = head; p2 = p2->next; } head = head->next; } p1->next = node2.next; p2->next = NULL; return node1.next; }};
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