Leetcode-14- longest common prefix (simple)

14 The longest common prefix （ Simple ）

Write a function to find the longest common prefix in the string array .

If no common prefix exists , Returns an empty string "".

Example 1：

 Input ：strs = ["flower","flow","flight"]
 Output ："fl"

Example 2：

 Input ：strs = ["dog","racecar","car"]
 Output ：""
 explain ： Input does not have a common prefix .

1. When the input is empty , non-existent strs[0], therefore , If the input is blank, it needs to be discussed separately
2. The longest public prefix length is not greater than strs Shortest string length in

（1）C++

class Solution {
public:
    string longestCommonPrefix(vector<string>& strs){
        int n = strs.size();
        if(n==0)
            return "";
        string s = "";
        string temp = strs[0];
        for(int i =1; i<n; i++){
            if(temp.length()<strs[i].length())
                temp = strs[i];
        }
        for(int j=0;j<temp.size();j++){
            char temp = strs[0][j];
            bool flag = true;
            for(int i=1; i<n ;i++){
                if(temp != strs[i][j])
                    flag = false;
            }
            if(flag)
                s+=temp;
            else
                break;
        }
            return s;
        }
};

(2)C++(C++ Boundary detection of non array strings in ）

class Solution {
public:
    string longestCommonPrefix(vector<string>& strs) {
        if(strs.size() == 0)
            return "";
        string s = strs[0];
        for(int i = 1; i < strs.size(); i++) {
            for(int j = 0; j < s.length(); j++) {
                if(s[j] != strs[i][j]) {
                    s = s.substr(0, j);
                    break;
                }
            }
        }
        return s;
    }
};

(3)python

class Solution:
    def longestCommonPrefix(self, strs: List[str]) -> str:
        if(len(strs)==0):
            return ""
        res = strs[0]
        for each in strs:
            if len(res)>len(each):
                res = each
        for i in range(len(strs)):
            for j in range(len(res)):
                if res[j]!= strs[i][j]:
                    res = res[:j]
                    break
        return res