【剑指 offer】删除链表中重复的节点

题目描述：

在一个排序的链表中，存在重复的结点，请删除该链表中重复的结点，重复的结点不保留，返回链表头指针。 例如，链表1->2->3->3->4->4->5 处理后为 1->2->5。

思路：

先写了一个非递归版，对链表遍历一遍可以解决。先找到头节点，然后对当前节点和前一个、后一个节点的值比较，如果不同，可以加入。

写完发现。嚯，别人家的递归就是优雅。

非递归版：

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

class Solution {
public:
    ListNode* deleteDuplication(ListNode* pHead) {
        if (pHead == NULL || pHead->next == NULL) return pHead;
        ListNode* st = NULL;

        if (pHead->val != pHead->next->val)
            st = pHead;
        else {
            ListNode* pre = pHead;
            pHead = pHead->next;
            while(st == NULL && pHead != NULL) {
                if (pHead->next == NULL) {
                    if (pHead->val != pre->val) {
                        st = pHead;
                        break;
                    }
                }
                if (pre->val != pHead->val && pHead->val != pHead->next->val) {
                    st = pHead;
                    break;
                }
                pre = pHead;
                pHead = pHead->next;
            }
        }

        if (st == NULL) return NULL;

        ListNode* pre = pHead;
        pHead = pHead->next;
        ListNode* ans = st;

        while(pHead != NULL) {
            if (pHead->next == NULL) {
                if (pre->val != pHead->val) {
                    st->next = pHead;
                    st = st->next;
                    st->next = NULL;
                }
            }
            else if (pre->val != pHead->val && pHead->val != pHead->next->val) {
                st->next = pHead;
                st = st->next;
            }
            pre = pHead;
            pHead = pHead->next;
        }
        st->next = NULL;
        return ans;
    }
};

递归版：

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

class Solution {
public:
    ListNode* deleteDuplication(ListNode* pHead) {
        if (pHead == NULL || pHead->next == NULL) return pHead;
        if (pHead->val == pHead->next->val) {
            ListNode* cur = pHead;
            while(cur != NULL && cur->val == pHead->val) {
                cur = cur->next;
            }
            return deleteDuplication(cur);
        }
        else {
            pHead->next = deleteDuplication(pHead->next);
        }
        return pHead;
    }
};