Remove Duplicates from Unsorted List

Source

Write a removeDuplicates() function which takes a list and deletes
any duplicate nodes from the list. The list is not sorted.

For example if the linked list is 12->11->12->21->41->43->21,
then removeDuplicates() should convert the list to 12->11->21->41->43.

If temporary buffer is not allowed, how to solve it?

题解1 - 两重循环

Remove Duplicates 系列题，之前都是已排序链表，这个题为未排序链表。

最容易想到的简单办法就是两重循环删除重复节点了，当前遍历节点作为第一重循环，当前节点的下一节点作为第二重循环。

C++

/**
 * Definition of ListNode
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *         this->val = val;
 *         this->next = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @param head: The first node of linked list.
     * @return: head node
     */
    ListNode *deleteDuplicates(ListNode *head) {
        if (head == NULL) return NULL;

        ListNode *curr = head;
        while (curr != NULL) {
            ListNode *inner = curr;
            while (inner->next != NULL) {
                if (inner->next->val == curr->val) {
                    inner->next = inner->next->next;
                } else {
                    inner = inner->next;
                }
            }
            curr = curr->next;
        }

        return head;
    }
};

Java

/**
 * Definition for ListNode
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param ListNode head is the head of the linked list
     * @return: ListNode head of linked list
     */
    public static ListNode deleteDuplicates(ListNode head) {
        if (head == null) return null;

        ListNode curr = head;
        while (curr != null) {
            ListNode inner = curr;
            while (inner.next != null) {
                if (inner.next.val == curr.val) {
                    inner.next = inner.next.next;
                } else {
                    inner = inner.next;
                }
            }
            curr = curr.next;
        }

        return head;
    }
}

源码分析

删除链表的操作一般判断node.next较为合适，循环时注意inner = inner.next和inner.next = inner.next.next的区别即可。

复杂度分析

两重循环，时间复杂度为 O(n^2), 空间复杂度近似为 O(1).

题解2 - 万能的 hashtable

使用辅助空间哈希表，节点值作为键，布尔值作为相应的值（是否为布尔值其实无所谓，关键是键）。

C++

/**
 * Definition of ListNode
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *         this->val = val;
 *         this->next = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @param head: The first node of linked list.
     * @return: head node
     */
    ListNode *deleteDuplicates(ListNode *head) {
        if (head == NULL) return NULL;

        // C++ 11 use unordered_map
        // unordered_map<int, bool> hash;
        map<int, bool> hash;
        hash[head->val] = true;
        ListNode *curr = head;
        while (curr->next != NULL) {
            if (hash.find(curr->next->val) != hash.end()) {
                ListNode *temp = curr->next;
                curr->next = curr->next->next;
                delete temp;
            } else {
                hash[curr->next->val] = true;
                curr = curr->next;
            }
        }

        return head;
    }
};

Java

/**
 * Definition for ListNode
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param ListNode head is the head of the linked list
     * @return: ListNode head of linked list
     */
    public static ListNode deleteDuplicates(ListNode head) {
        if (head == null) return null;

        ListNode curr = head;
        HashMap<Integer, Boolean> hash = new HashMap<Integer, Boolean>();
        hash.put(curr.val, true);
        while (curr.next != null) {
            if (hash.containsKey(curr.next.val)) {
                curr.next = curr.next.next;
            } else {
                hash.put(curr.next.val, true);
                curr = curr.next;
            }
        }

        return head;
    }
}

源码分析

删除链表中某个节点的经典模板在while循环中体现。

复杂度分析

遍历一次链表，时间复杂度为 O(n), 使用了额外的哈希表，空间复杂度近似为 O(n).