解题思路:
(1)先求出最短的字符串长度
(2)再分别判断每个字符串对应位置是否相等
class Solution {
public:
string longestCommonPrefix(vector<string>& strs) {
if (strs.size()==0) return "";
int length=INT_MAX,j=0,tag=-1;
for (auto w:strs) {
if (length>w.length()) length = w.length();
}
for (int i=0;i<length;i++) {
j=0;
while(j<strs.size()-1) {
if(strs[j][i]==strs[j+1][i]) {
j++;
} else {
tag=i;
break;
}
}
if (tag!=-1) {
if(tag==0) return "";
else break;
}
}
if (tag==-1) tag=length;
return strs[0].substr(0,tag);
}
};