Write a function to find the longest common prefix string amongst an array of strings. 思路:找最长公共前缀 常规方法 string longestCommonPrefix(vector<string> &strs) { ) return ""; ) ]; string ans; ; ) { ; i < strs.size(); i++) { ].size() <= n…
题目简述: Write a function to find the longest common prefix string amongst an array of strings. 解题思路: class Solution: # @return a string def longestCommonPrefix(self, strs): if strs == []: return '' minl = 99999 for i in strs: if len(i) < minl: minl = l…
Longest Common Prefix Write a function to find the longest common prefix string amongst an array of strings. 题目的意图 It seems that it is not to check between pair of strings but on all the strings in the array. For example: {"a","a",&quo…
这道题是LeetCode里的第14道题. 题目描述: 编写一个函数来查找字符串数组中的最长公共前缀. 如果不存在公共前缀,返回空字符串 "". 示例 1: 输入: ["flower","flow","flight"] 输出: "fl" 示例 2: 输入: ["dog","racecar","car"] 输出: "" 解释: 输入…
Reverse a singly linked list. 思路:没啥好说的.秒... ListNode* reverseList(ListNode* head) { ListNode * rList = NULL, * tmp = NULL; while(head != NULL) { tmp = rList; rList = head; head = head->next; rList->next = tmp; } return rList; }…
The count-and-say sequence is the sequence of integers beginning as follows:1, 11, 21, 1211, 111221, ... 1 is read off as "one 1" or 11.11 is read off as "two 1s" or 21.21 is read off as "one 2, then one 1" or 1211. Given an…
Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. 思路:编程之美里有,就是找因子5的个数. int trailingZeroes(int n) { ; ) { ans += n / ; n /= ; } return ans; }…
这题实现起来还是挺麻烦的,就偷懒使用下库函数strtod().第二个参数表示字符中不是数字的地方,如果后面是空格,则认为其仍是数字,否则不是. bool isNumber(char *s) { char *endptr; strtod(s, &endptr); if (endptr == s)return false; for (; endptr; endptr++) { if (!isspace(*endptr)) return false; } return true; }…
在一组字符串中找到最长的子串.采用纵向匹配,遇到第一个不匹配的停止. string longestComPrefix(vector<string> &strs) { if (strs.empty())return " "; //纵向比较 ; idx < strs[].size(); idx++) { ; i < strs.size(); i++) { ][idx] != strs[i][idx])].substr(, idx); } } ]; }…
[SPOJ]Longest Common Substring(后缀自动机) 题面 Vjudge 题意:求两个串的最长公共子串 题解 \(SA\)的做法很简单 不再赘述 对于一个串构建\(SAM\) 另外一个串在\(SAM\)上不断匹配 最后计算答案就好了 匹配方法: 如果\(trans(s,c)\)存在 直接沿着\(trans\)走就行,同时\(cnt++\) 否则沿着\(parent\)往上跳 如果存在\(trans(now,c),cnt=now.longest+1\) 否则,如果不存在可行的…