LeetCode - 86 - Partition List
2017-07-20 15:39
501 查看
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
return
居然越来越喜欢链表了,说不出自己都不敢相信【笑哭】也渐渐开始区分 -> 和 . 用在哪里了
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;
}
};
You should preserve the original relative order of the nodes in each of the two partitions.
For example,
Given
1->4->3->2->5->2and x = 3,
return
1->2->2->4->3->5.
居然越来越喜欢链表了,说不出自己都不敢相信【笑哭】也渐渐开始区分 -> 和 . 用在哪里了
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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