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[Leetcode]147. Insertion Sort List(C++)

题目描述

题目链接:147. Insertion Sort List

Given the head of a singly linked list, sort the list using insertion sort, and return the sorted list’s head.

The steps of the insertion sort algorithm:

  1. Insertion sort iterates, consuming one input element each repetition and growing a sorted output list.
  2. At each iteration, insertion sort removes one element from the input data, finds the location it belongs within the sorted list and inserts it there.
  3. It repeats until no input elements remain.
    The following is a graphical example of the insertion sort algorithm. The partially sorted list (black) initially contains only the first element in the list. One element (red) is removed from the input data and inserted in-place into the sorted list with each iteration.

例子

例子 1

Input: head = [4,2,1,3]
Output: [1,2,3,4]

例子 2

Input: head = [-1,5,3,4,0]
Output: [-1,0,3,4,5]

Constraints

  • The number of nodes in the list is in the range [1, 5000].
  • -5000 <= Node.val <= 5000

解题思路

这道题题目已经给出解题方法了,用一个新节点储存新链表,然后遍历原链表,对原链表的每一个节点在新链表中遍历一次找到合适位置插入即可,代码如下:

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/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
#include <limits.h>

class Solution {
public:
ListNode* insertionSortList(ListNode* head) {
ListNode* dummy = new ListNode(INT_MIN);
while (head) {
ListNode* next = head->next;
insertNode(dummy, head);
head = next;
}

return dummy->next;
}

private:
void insertNode(ListNode* head, ListNode* node) {
if (!head) return head = node;
ListNode* prev = head;
ListNode* curr = head->next;
while (curr && curr->val < node) {
prev = curr;
curr = curr->next;
}
prev->next = node;
node->next = curr;
}
};
  • 时间复杂度: O(n^2)
  • 空间复杂度: O(1)

GitHub 代码同步地址: 147.InsertionSortList.cpp

其他题目:
GitHub: Leetcode-C++-Solution
博客: Leetcode-Solutions