leetcode-14最长公共前缀(分治|二分)[通俗易懂]

示例 1：

输入：strs = ["flower","flow","flight"]
输出："fl"
示例 2：

输入：strs = ["dog","racecar","car"]
输出：""
解释：输入不存在公共前缀。

class Solution { 
   
public:
    string com(string a,string b){ 
   
        int len = min(a.size(),b.size());
        int i = 0;
        while(i < len && a[i] == b[i])i ++;
        return a.substr(0,i);
    }
    string div(vector<string> &strs,int l,int r){ 
   
        if(l < r){ 
   
            int mid = l + r >> 1;
            string left = div(strs,l,mid);
            string right = div(strs,mid + 1,r);
            int len = min(left.size(),right.size());
            int i = 0;
            while(i < len && left[i] == right[i])i ++;
            return left.substr(0,i);
        }else return strs[l];
    }
    string longestCommonPrefix(vector<string>& strs) { 
   
        if(!strs.size())return "";
        return div(strs,0,strs.size() - 1);
    }
};

class Solution { 
   
public:
    bool check(vector<string>strs,int x){ 
   

        for(int i = 0;i <= x;i ++){ 
   
            for(int j = 1;j < strs.size();j ++){ 
   
                if(strs[j][i] != strs[0][i])return false;
            }
        }
        return true;
    }
    int min(int x,int y){ 
   
        return x < y ? x : y;
    }
    string longestCommonPrefix(vector<string>& strs) { 
   
        if(!strs.size())return "";
        int mlen = strs[0].size();
        for(int i = 1;i < strs.size();i ++)mlen = min(mlen,strs[i].size());
        int l = -1,r = mlen - 1;
        while(l < r){ 
   
            int mid = (l + r + 1)>> 1;
            if(check(strs,mid))l = mid;
            else r = mid - 1;
        }
        return strs[0].substr(0,l + 1);
    }
};