[LeetCode] Insertion Sort List

Sort a linked list using insertion sort.

解题思路

对于得到结点current的插入位置，从头结点开始遍历，直到遍历到值大于等于节点current的结点，然后将从该结点到current的前驱结点的所有结点的值依次和current结点的值交换，从而达到将该节点插入所遍历到的位置的目的。

实现代码

/*****************************************************************
*  @Author   : 楚兴
*  @Date     : 2015/2/16 00:06
*  @Status   : Accepted
*  @Runtime  : 287 ms
******************************************************************/
#include <iostream>

using namespace std;

struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}

};

class Solution {
public:
ListNode *insertionSortList(ListNode *head) {
if (head == NULL || head->next == NULL)
{
return head;
}
ListNode *current = head->next;
ListNode *temp;
while(current != NULL)
{
temp = head;
while(temp != current && temp->val < current->val)
{
temp = temp->next;
}
if (temp != current)
{
while (temp != current)
{
swap(temp->val, current->val);
temp = temp->next;
}
}
current = current->next;
}
return head;
}
};

void print(ListNode *head)
{
ListNode *tmp = head;
while(tmp)
{
cout<<tmp->val<<" ";
tmp = tmp->next;
}
cout<<endl;
}

int main()
{
ListNode* head = new ListNode(12);
ListNode* node1 = new ListNode(8);
head->next = node1;
ListNode* node2 = new ListNode(5);
node1->next = node2;
ListNode* node3 = new ListNode(9);
node2->next = node3;
ListNode* node4 = new ListNode(0);
node3->next = node4;
ListNode* node5 = new ListNode(1);
node4->next = node5;
ListNode* node6 = new ListNode(3);
node5->next = node6;
ListNode* node7= new ListNode(6);
node6->next = node7;
ListNode* node8 = new ListNode(4);
node7->next = node8;
print(head);

Solution s;
s.insertionSortList(head);
print(head);
}

上述代码虽然已经被AC，但是用时还是太久，感觉主要是因为交换的次数太多了。故将代码更改如下：

/*****************************************************************
*  @Author   : 楚兴
*  @Date     : 2015/2/16 14:44
*  @Status   : Accepted
*  @Runtime  : 96 ms
******************************************************************/
class Solution {
public:
ListNode *insertionSortList(ListNode *head) {
if (head == NULL || head->next == NULL)
{
return head;
}

ListNode *cur = head->next;  //当前结点
ListNode *pre = head;  //当前结点的前驱结点
ListNode *temp;
ListNode *newcur;

while(cur != NULL)
{
temp = head;
newcur = cur->next;
if (temp == head && temp->val > cur->val)  //头结点的值大于当前结点的值
{
pre->next = pre->next->next;
cur->next = temp;
head = cur;
cur = newcur;
continue;
}

//找到其后继结点的值不大于不大于当前结点值的temp结点
while (temp->next != cur && temp->next->val < cur->val)
{
temp = temp->next;
}

if (temp->next != cur)
{
pre->next = pre->next->next;

ListNode *t = temp->next;
temp->next = cur;
cur->next = t;
}
else
{
//位于当前结点之前的所有结点的值均小于当前结点
pre = pre->next;
}
cur = newcur;
}

return head;
}
};