83. Remove Duplicates from Sorted List [LeetCode]_c.set operations automatically eliminate duplicate-优快云博客

本文介绍三种方法来删除已排序链表中的重复元素，确保每个元素仅出现一次。第一种方法为迭代方式，通过指针遍历链表并移除重复节点；第二种和第三种方法采用递归方式实现，分别提供了不同的递归逻辑来完成任务。

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83. Remove Duplicates from Sorted List

Given a sorted linked list, delete all duplicates such that each element appear only once.

Example 1:

Input: 1->1->2
Output: 1->2

Example 2:

Input: 1->1->2->3->3
Output: 1->2->3

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */

typedef struct ListNode NODE;

///
///
/// Approach 1: Straight-Forward Approach
NODE *deleteDuplicates(NODE *head){
    if (NULL == head || NULL == head->next) return head;
    
    NODE *pre = head;
    NODE *cur = head->next;
    while (cur) {
        if (pre->val == cur->val) {
            pre->next = cur->next;
            free(cur);
        } else {
            pre = cur;
        }
        cur = pre->next;
    }
    return head;
}

///
///
/// Approach 2: recursion version
NODE *deleteDuplicates(NODE *head){
    if (NULL == head || NULL == head->next) return head;
    NODE *next = head->next;
    if (head->val == next->val) {
        while (next && head->val == next->val) {
            //NODE *del = next;
            next = next->next;
            //free(del);
        }
        head->next = deleteDuplicates(next); 
    } else {
        head->next = deleteDuplicates(head->next);
    }
    return head;
}

///
///
/// Approach 3: another recursion version
static void recur(NODE *pre, NODE *cur) {
    if (NULL == cur) return;
    if (pre->val == cur->val) {
        pre->next = cur->next;
        free(cur);
        recur(pre, pre->next);
    } else {
        recur(pre->next, cur->next);
    }
}

NODE *deleteDuplicates(NODE *head){
    if (NULL == head || NULL == head->next) return head;
    NODE dum;
    dum.val = head->val + 1;
    dum.next = head;
    recur(&dum, head);
    return dum.next;
}