所有字符串的公共前缀最长字符串
特点:(1)公共所有字符串前缀 (好像跟没说一样。。。)
(2)在字典树中特点:任意从根节点触发遇见第一个分支为止的字符集合即为目标串
参考问题:https://leetcode.com/problems/longest-common-prefix/description/
Write a function to find the longest common prefix string amongst an array of strings. If there is no common prefix, return an empty string "". Example 1: Input: ["flower","flow","flight"] Output: "fl" Example 2: Input: ["dog","racecar","car"] Output: "" Explanation: There is no common prefix among the input strings. Note: All given inputs are in lowercase letters a-z.
题干容易理解,翻译略
实现步骤
1.构建字典树
当任意一字符串中的当前节点的branchCount等于输入的单词个数时候,那么这个节点就是在最长前缀里
C 语言解法:
#define MAX 30 //the total number of alphabet is 26, a...z struct DicTrie{ bool isTerminal;//是否是单词结束标志 int count; //当前字符串出现次数 int branchCount; //计数当前节点的孩子数 struct DicTrie *next[MAX ]; //每个节点 最多 有 MAX 个孩子节点 结构体嵌套 }; int insertTrie(struct DicTrie *root ,char *targetString) { if (!targetString) { return 0; } int len = strlen(targetString); if (len <= 0) { return 0; } struct DicTrie *head = root; for (int i = 0; i < len; i ++) { int res = (int)(targetString[i] - 'a');//当前小写字母对应数字 if (head->next[res] == NULL) { //如果是空节点 head->next[res] = (struct DicTrie *)malloc(sizeof(struct DicTrie));//new DicTrie;//则插入新节点元素 head = head->next[res]; //更新头指针 并初始化 head->count = 0; // for (int j = 0; j < MAX; j ++) { head->next[j] = NULL; head->isTerminal = false; } head->branchCount = 1;//一个分支 } else { head = head->next[res]; head->branchCount ++;//分支累计 } } head->count ++;//每次插入一个,响应计数都增加1 head->isTerminal = true; return head->count; } char* longestCommonPrefix(char** strs, int strsSize) { int len = strsSize; //边界处理 if (len == 0) { return ""; } if (len == 1) { return strs[0]; } //组织字典树 struct DicTrie *root = NULL; root = (struct DicTrie *)malloc(sizeof(struct DicTrie)); root->count = 0; root->branchCount = 0; for (int i = 0; i < MAX; i ++) { root->next[i] = NULL; // 空节点 root->isTerminal = false; // } // for (int i = 0;i < len; i ++) { insertTrie(root, strs[i]); } // int preIndex = 0; struct DicTrie *head = root; bool isFlag = false; int i = 0; int count = strlen(strs[0]);//任意一字符串都可以 从strs[0]中查即可 for (preIndex = 0; preIndex< count; preIndex ++) { int targetIndex = strs[0][preIndex] - 'a'; head = head->next[targetIndex]; if (head->branchCount == len) { i ++;//拿到合法前缀的计数 isFlag = true; } } if (isFlag) { preIndex = i; } else { preIndex = 0; } strs[0][preIndex] = '\0'; return strs[0]; }
自己编辑时候的主函数:
int main(int argc, const char * argv[]) { // insert code here... char *s[30]= {"dog","dracecar","dcar"}; // char *s[30]= {"flower","flow","flight"}; char *str = longestCommonPrefix(s,3); printf("%s",str); return 0; }
其实,我这道题在思路上没有任何问题,工作用的都是面向对象语言,面向过程C,纯C少了,所以代码不符合提交要求
比如创建结构体 我自己写 就是new DicTrie,但是纯C种 用malloc.
还有字符串截取。。。都得自己一点点面向过程敲。不然就是运行错误。
最后
(1)解这道题的目的:
(2)拓宽思路后缀数组:
(3)这道题的高效解法:
该休息了,剩下的后续补充