[LeetCode]77. Reverse Linked List反轉鏈表

Reverse a singly linked list.

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Hint:

A linked list can be reversed either iteratively or recursively. Could you implement both?

 

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解法1：一個最簡單的辦法就是借助棧的后進先出功能，先掃描一遍鏈表保存每個節點的值，然后再從頭到尾遍歷，將棧中元素值一一賦給鏈表節點。時空復雜度都是O(n)。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        stack<int> elem;
        ListNode* curr = head;
        while(curr != NULL) {
            elem.push(curr->val);
            curr = curr->next;
        }
        curr = head;
        while(curr != NULL) {
            curr->val = elem.top();
            curr = curr->next;
            elem.pop();
        }
        return head;
    }
};

 

解法2：可以做到in-place的反轉。鏈表反轉后，實際上只是中間節點的指針反轉，並且反轉后原來鏈表的頭結點的下一個節點應該為NULL，而反轉后鏈表的頭結點為原來鏈表的尾節點。我們可以從頭結點開始，每次處理兩個節點之間的一個指針，將其反轉過來。然后再處理接下來兩個節點之間的指針……直至遇到尾節點，設置為新鏈表的頭結點即可。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode* rHead = NULL; // 反轉后的頭節點
        ListNode* curr = head; // 當前處理節點
        ListNode* pTail = NULL; // 反轉后尾節點
        while(curr != NULL) {
            ListNode* pNext = curr->next;
            if(pNext == NULL)
                rHead = curr;
            curr->next = pTail;
            pTail = curr;
            curr = pNext;
        }
        return rHead;
    }
};

上面的是一個循環來進行反轉。遞歸的方式如下：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if (head == NULL || head->next == NULL) return head;
        ListNode* rHead = reverseList(head->next); // 反轉得到新鏈表的頭節點
        head->next->next = head; // 當前節點的下一個節點的next指針反轉過來
        head->next = NULL; // 設置新鏈表的尾節點
        return rHead;
    }
};