【leetcode】14. Longest Common Prefix(Python & C++)
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14. Longest Common Prefix
题目链接
14.1 题目描述:
Write a function to find the longest common prefix string amongst an array of strings.
14.2 解题思路:
- 思路一:思想是以strs中第一字符串为模板,每次比较其字符与其他字符串相同位置的字符,并且判断没有超过字符串的长度。初始化公共前缀pre=”“。如果strs为空,直接返回pre。外循环,以strs中第一个字符串的长度遍历,初始i=0。如果i大于等于strs[0].length(),则返回pre。初始化每次比较的字符temp=strs[0][i]。内循环strs中的每个字符串。判断,如果i 大于等于 strs[j].length()或者strs[j][i]与temp的值不相等,则返回pre。内循环结束,则将temp添加到pre尾部。外循环结束,返回pre。
14.3 C++代码:
1、思路一代码(6ms):
class Solution117 {public: string longestCommonPrefix(vector<string>& strs) { string pre = ""; if (strs.size() == 0) return pre; for (int i = 0; i < strs[0].length();i++) { if (i >= strs[0].length()) return pre; char temp = strs[0][i]; for (int j = 0; j < strs.size();j++) { if (i >= strs[j].length() || strs[j][i] != temp) return pre; } pre.append(1,temp); } return pre; }};
14.4 Python代码:
1、思路一代码(49ms)
class Solution(object): def longestCommonPrefix(self, strs): """ :type strs: List[str] :rtype: str """ pre="" if len(strs)==0: return pre for i in range(len(strs[0])): if i>=len(strs[0]): return pre temp=strs[0][i] for j in range(len(strs)): if i>=len(strs[j]) or strs[j][i]!=temp: return pre pre=pre+temp return pre
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