力扣 86. 分隔链表
题目描述
给你一个链表和一个特定值 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。
你应当保留两个分区中每个节点的初始相对位置。
示例:
输入:head = 1->4->3->2->5->2, x = 3
输出:1->2->2->4->3->5
解决方案
方法:模拟
直观来说我们只需维护两个链表 small 和 large 即可,small 链表按顺序存储所有小于 x 的节点,large 链表按顺序存储所有大于等于 x 的节点。遍历完原链表后,我们只要将 small 链表尾节点指向 large 链表的头节点即能完成对链表的分隔。
为了实现上述思路,我们设 smallHead 和 largeHead 分别为两个链表的哑节点,即它们的 next 指针指向链表的头节点,这样做的目的是为了更方便地处理头节点为空的边界条件。同时设 small 和 large 节点指向当前链表的末尾节点。开始时 smallHead = small,largeHead = large。随后,从前往后遍历链表,判断当前链表的节点值是否小于 x,如果小于就将 small 的 next 指针指向该节点,否则将 large 的 next 指针指向该节点。
遍历结束后,我们将 large 的 next 指针置空,这是因为当前节点复用的是原链表的节点,而其 next 指针可能指向一个小于 x 的节点,我们需要切断这个引用。同时将 small 的 next 指针指向 largeHead 的 next 指针指向的节点,即真正意义上的 large 链表的头节点。最后返回 smallHead 的 next 指针即为我们要求的答案。
代码
C++
class Solution {
public:
ListNode* partition(ListNode* head, int x) {
ListNode* small = new ListNode(0);
ListNode* smallHead = small;
ListNode* large = new ListNode(0);
ListNode* largeHead = large;
while (head != nullptr) {
if (head->val < x) {
small->next = head;
small = small->next;
} else {
large->next = head;
large = large->next;
}
head = head->next;
}
large->next = nullptr;
small->next = largeHead->next;
return smallHead->next;
}
};
Java
class Solution {
public ListNode partition(ListNode head, int x) {
ListNode small = new ListNode(0);
ListNode smallHead = small;
ListNode large = new ListNode(0);
ListNode largeHead = large;
while (head != null) {
if (head.val < x) {
small.next = head;
small = small.next;
} else {
large.next = head;
large = large.next;
}
head = head.next;
}
large.next = null;
small.next = largeHead.next;
return smallHead.next;
}
}
JavaScript
var partition = function(head, x) {
let small = new ListNode(0);
const smallHead = small;
let large = new ListNode(0);
const largeHead = large;
while (head !== null) {
if (head.val < x) {
small.next = head;
small = small.next;
} else {
large.next = head;
large = large.next;
}
head = head.next;
}
large.next = null;
small.next = largeHead.next;
return smallHead.next;
};
Golang
func partition(head *ListNode, x int) *ListNode {
small := &ListNode{}
smallHead := small
large := &ListNode{}
largeHead := large
for head != nil {
if head.Val < x {
small.Next = head
small = small.Next
} else {
large.Next = head
large = large.Next
}
head = head.Next
}
large.Next = nil
small.Next = largeHead.Next
return smallHead.Next
}
C
struct ListNode* partition(struct ListNode* head, int x) {
struct ListNode* small = malloc(sizeof(struct ListNode));
struct ListNode* smallHead = small;
struct ListNode* large = malloc(sizeof(struct ListNode));
struct ListNode* largeHead = large;
while (head != NULL) {
if (head->val < x) {
small->next = head;
small = small->next;
} else {
large->next = head;
large = large->next;
}
head = head->next;
}
large->next = NULL;
small->next = largeHead->next;
return smallHead->next;
}
复杂度分析
时间复杂度: O(n),其中 n 是原链表的长度。我们对该链表进行了一次遍历。
空间复杂度: O(1)。
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