Write a function to find the longest common prefix string amongst an array of strings.
题意:找出所给几个字符串的相同前缀
思路:用第一个字符串和之后的所有字符串进行对比,标示出相同字符串的超尾指针就行
ps:通过这个题发现自己Python基础好差啊
1 class Solution(object): 2 def longestCommonPrefix(self, strs): 3 """ 4 :type strs: List[str] 5 :rtype: str 6 """ 7 if len(strs)==0: return '' 8 str = strs[0] 9 min = len(str) 10 for i in range(1, len(strs)): 11 j=0 12 tmp = strs[i] 13 while j<min and j<len(tmp) and str[j] == tmp[j]: 14 j+=1 15 if j<min: min = j 16 17 return str[:min]
耗时84ms,排名太靠后了,不能忍
1 class Solution(object): 2 def longestCommonPrefix(self, strs): 3 """ 4 :type strs: List[str] 5 :rtype: str 6 """ 7 if len(strs)==0: return '' 8 str = strs[0] 9 min = len(str) 10 for tmp in strs[1:]: 11 j=0 12 while j<min and j<len(tmp) and str[j]==tmp[j]: 13 j+=1 14 if j<min: min=j 15 16 return str[:min]
耗时59ms,好了点,但还是不理想
1 class Solution(object): 2 def longestCommonPrefix(self, strs): 3 """ 4 :type strs: List[str] 5 :rtype: str 6 """ 7 if len(strs)==0: return '' 8 str = strs[0] 9 min = len(str) 10 for tmp in strs[1:]: 11 j = 0 12 l = len(tmp) 13 while j<min and j<l and str[j]==tmp[j]: 14 j+=1 15 if j<min: min=j 16 17 return str[:min]
把重复计算的min = len(str)保存后,变成了49ms,len()耗时挺长啊,感觉还能继续改
1 class Solution(object): 2 def longestCommonPrefix(self, strs): 3 """ 4 :type strs: List[str] 5 :rtype: str 6 """ 7 if len(strs)==0: return '' 8 str = strs[0] 9 min = len(str) 10 for tmp in strs[1:]: 11 j = 0 12 l = len(tmp) 13 while j<min and j<l and str[j]==tmp[j]: 14 j+=1 15 min=j #不用判断,j的值肯定小于min,直接更新 16 17 return str[:min]
又减了4ms
感觉还能改,但是不知道怎么改了。。。
没办法,
只有这样了。。
!!!!
0ms
1 char* longestCommonPrefix(char** strs, int strsSize) { 2 if(!strsSize) return ""; 3 char *str=strs[0]; 4 int i,j; 5 for(i=1;i<strsSize;i++){ 6 j=0; 7 while(str[j]&&strs[i][j]&&str[j]==strs[i][j]) 8 j++; 9 str[j]='