Leetcode 题目整理-4 Longest Common Prefix & Remove Nth Node From End of List

14. Longest Common Prefix

Write a function to find the longest common prefix string amongst an array of strings.

注：这题竟然连个示例都没有，说明特殊情况并不多，就是要找出所有字符串的最长公共前缀。他应该不会超过所有字符串中最短的那个，可以试着找出最短长度n，然后每个都读取和比较n个，并根据情况不断减小那个n.\

 if (strs.size() == )
    {
        return "";
    }
    int n{int(strs.front().size())};
    if (n == )
    {
        return "";
    }
    for (vector<string>::iterator str_i = strs.begin()+; str_i != strs.end(); str_i++)
    {
        if ((*str_i).size() < n)
        {
            n = (*str_i).size();
        }
        int temp{n};//感觉有点儿多余
        for (int i = ; i < n; i++)
        {
            if ((*str_i).at(i) != (*(str_i - )).at(i))
            {
                temp = i;
                break;
            }
        }
        n = temp;
    }

    string longest_prefix(strs.front(),,n);
    return longest_prefix;

19. Remove Nth Node From End of List

Given a linked list, remove the n^th node from the end of list and return its head.

For example,

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:
Given n will always be valid.
Try to do this in one pass.

注：给出一个链表，把倒数第n个元素移除。听起来很常用的技术，了解一下链表的结构，用处 ，优势，再来做这个题。

今天关于链表的部分没有完成，明天继续

    int i{  };
    ListNode *temp,*fronter ,*backer,*curNode;
    temp = head;
    //fronter从第一个节点开始向后移动，直到第n个，或者到了结尾
    for (fronter = temp; i < n && !(fronter->next==NULL); fronter = fronter->next,i++);//(*fronter).next)
    if (fronter->next == NULL)//如果是因为到了结尾而结束的循环，那么说明要删除的点在head
    {
        ListNode *tmp = head;
        if (head->next==NULL)//整个链表只有一个节点
            head= ;
        else
            head = head->next;
        return head;
    }
    //如果是因为fornter已经过去n个了，那么这个时候给backer赋值
    curNode = head;
    //二者同时继续向后遍历，直到fronter走到结尾时，删除backer之后的那个节点
    for (; fronter->next != NULL; fronter = fronter->next){

        backer=curNode;
        curNode = curNode->next;
    }
    //删除backer后面的节点
    backer->next = curNode->next;
    return head;