leetcode-19-Remove Nth Node From End of List

问题

题目：[leetcode-19]

思路

需要注意一点，当n==len的情形需要特殊处理。

代码

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* removeNthFromEnd(ListNode* head, int n) {        int len = 0;        for( ListNode* p = head; p!=NULL; p=p->next ) len++;        if(n==len)        {            ListNode* p = head;            head = head->next;            delete p;            p = NULL;            return head;        }        int cnt = len-n;        ListNode* pre = NULL;        ListNode* p = head;        for(int i = 0; i < cnt; ++i){             pre = p;            p=p->next;        }        pre->next = p->next;        delete p;        p = NULL;        return head;    }};

思路1

本省可以两个指针。但是边界情况太多。
所以，先算出节点个数了。
边界情况比较多，小心！

代码1

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* removeNthFromEnd(ListNode* head, int n) {        if(!head) return NULL; // 1        ListNode* p = head;        int cnt = 0;        while(p){            ++cnt;            p = p->next;        }        if(cnt < n) return NULL; // 2        if(1==cnt) return NULL; // 3        if(n==cnt){ // 4            ListNode* ret = head->next;            delete head;            return ret;        }        int k = cnt - n - 1;        std::cout << k << std::endl;        p = head;        for(int i = 0; i < k; ++i){            p = p->next;        }        ListNode* q = p->next;        p->next = q->next;        delete q;        return head;    }};