[LeetCode] 翻转子链表 reverse linked list

最新推荐文章于 2022-09-09 22:24:36 发布

jiyanfeng1

最新推荐文章于 2022-09-09 22:24:36 发布

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分类专栏：链表文章标签： leetcode

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本文介绍了一种在不使用额外空间的情况下，通过迭代方式反转链表中指定位置区间内的元素，实现链表部分反转的功能。

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Reverse a linked list from position m to n. Do it in-place and in one-pass. For example: Given 1->2->3->4->5->NULL, m = 2 and n = 4, return 1->4->3->2->5->NULL. Note: Given m, n satisfy the following condition: 1 ≤ m ≤ n ≤ length of list.

下面给出一个简单的迭代算法（自己的），要注意算法的逻辑。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *reverseBetween(ListNode *head, int m, int n) {
        
        if(m==n)
            return head;
            
        ListNode* prev=NULL;
        ListNode* curr=head;
        ListNode* next=NULL;
        
        // the head node is at position 1
        int pos=1;
        
        // find the node at position m
        while(pos<m)
        {
            prev = curr;
            curr = curr->next;
            pos++;
        }
        
        ListNode* old_prev = prev;
        ListNode* old_curr = curr;

        // start with the node at position m+1
        prev = curr;
        curr = curr->next;
		next = curr->next;
        pos++;
        
        while(pos<=n)
        {
            curr->next = prev;
            prev = curr;
            curr = next;
            if(curr)
                next = curr->next;
            pos++;
        }
        
        if(m>1)
            old_prev->next = prev;
            
        old_curr->next = curr;
        
        if(m>1)
            return head;
        else
            return prev;
    }
};

下面是一个更简洁的代码：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *reverseBetween(ListNode *head, int m, int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        ListNode* h=head;
        for (int i=0;i<n-m;i++){
            int k1=m+i;
            int k2=n-i;
            if (k1>=k2){return head;}
            ListNode* p = h;
            ListNode* q = h;
            while (k1-1>0){p=p->next;k1--;}
            while (k2-1>0){q=q->next;k2--;}
            int tmp=p->val;
            p->val=q->val;
            q->val=tmp;
        }
        return head;
    }
};