Leetcode 14 最长公共前缀

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 “”。

题解1(字符串排序)

class Solution {
public:
    string longestCommonPrefix(vector<string>& strs) {
        if(strs.empty())
            return string();
         //利用字符串排序
        sort(strs.begin(), strs.end());
        //随便挑两个
        string f_str = strs.front();
        string e_str = strs.back();
        size_t i = 0;
        while(i < min(f_str.size(), e_str.size())){
            if(f_str[i] == e_str[i])
                i++;
            else
                break;
        }
        return f_str.substr(0,i);

    }
};

题解2(横向比较)

class Solution {
public:
    string longestCommonPrefix(vector<string>& strs) {
        if(strs.empty())
            return string();
        size_t s_str = strs.size();
        string fianls = strs[0];
        for(size_t i = 1; i < s_str; i++){
            size_t j = 0;
            while(j < min(fianls.size(), strs[i].size())){
                if(fianls[j] == strs[i][j])
                    j++;
                else break;
            }
            fianls = fianls.substr(0,j);
            if(! fianls.size())
                return string();
        }
        return fianls;
    }
};

题解3(纵向比较，注意越界)

class Solution {
public:
    string longestCommonPrefix(vector<string>& strs) {
        if(strs.empty())
            return string();
        size_t s_str = strs.size();
        string finals = "";
        for(size_t i = 0; i < strs[0].size();){
            size_t j = 0;
            while(j < s_str-1){
                if(i == strs[j].size())
                    return finals;
                if(strs[j][i] == strs[j+1][i])
                    j++;
                else break;
            }
            if(j == s_str-1){
                finals += strs[0][i];
                i++;
            }else{
                return finals;
            }
        }
        return finals;
    }
};

题解4(大神的排序代码)

class Solution {
public:
    string longestCommonPrefix(vector<string>& strs) {
        if(strs.empty()) return "";
        // c++17 结构化绑定
        // str0, str1 分别是一个 pair 的 first 和 second
        const auto [str0, str1] = minmax_element(strs.begin(), strs.end());
        for(int i = 0; i < str0->size(); ++i)
            if(str0->at(i) != str1->at(i)) return str0->substr(0, i);
        return *str0;
    }
};