Write a function to find the longest common prefix string amongst an array of strings.
“abcdefg”
“abcdefghijk”
“abcdfghijk”
“abcef”
上面的字符串数组的最长公共前缀就是”abc”。
class Solution {
public:
string longestCommonPrefix(vector<string>& strs) {
if(strs.size() <= 0)
return "";
sort(strs.begin(),strs.end());
string a = strs[0];
string b = strs[strs.size()-1];
string res;
for(int i = 0;i < a.length();++i)
{
if(a[i] == b[i])
res = res + a[i];
else
break;
}
return res;
}
};
算法1:逐个字符比较,时间复杂度为O(N*L),N是字符串个数,L是最长前缀的长度
class Solution {
public:
string longestCommonPrefix(vector<string> &strs) {
int n = strs.size();
string res;
if(n == 0)return res;
for(int pos = 0; pos < strs[0].size(); pos++)//最长前缀的长度不超过strs[0].size(),逐个字符的比较
{
for(int k = 1; k < n; k++)//strs[0]的第pos个字符分别和strs[1...n-1]的第pos个字符比较
{
if(strs[k].size() == pos || strs[k][pos] != strs[0][pos])
return res;
}
res.push_back(strs[0][pos]);
}
return res;
}
};
class Solution {
public:
string longestCommonPrefix(vector<string>& strs) {
if (strs.empty()) return "";
string res = "";
for (int j = 0; j < strs[0].size(); ++j) {
char c = strs[0][j];
for (int i = 1; i < strs.size(); ++i) {
if (j >= strs[i].size() || strs[i][j] != c) {
return res;
}
}
res.push_back(c);
}
return res;
}
};
我们可以对上面的方法进行适当精简,如果我们发现当前某个字符和下一行对应位置的字符不相等,说明不会再有更长的共同前缀了,我们直接通过用substr的方法直接取出共同前缀的子字符串。如果遍历结束前没有返回结果的话,说明第一个单词就是公共前缀,返回为结果即可。代码如下:
class Solution {
public:
string longestCommonPrefix(vector<string>& strs) {
if (strs.empty()) return "";
for (int j = 0; j < strs[0].size(); ++j) {
for (int i = 0; i < strs.size() - 1; ++i) {
if (j >= strs[i].size() || j >= strs[i + 1].size() || strs[i][j] != strs[i + 1][j]) {
return strs[i].substr(0, j);
}
}
}
return strs[0];
}
};