Given the array of strings A, you need to find the longest string S which is the prefix of ALL the strings in the array.
Longest common prefix for a pair of strings S1 and S2 is the longest string S which is the prefix of both S1 and S2.
For Example, longest common prefix of "abcdefgh" and "abcefgh" is "abc".
Input Format
The only argument given is an array of strings A.
Output Format
Return longest common prefix of all strings in A.
For Example
Input 1:
A = ["abcdefgh", "aefghijk", "abcefgh"]
Output 1:
"a"
Explanation 1:
Longest common prefix of all the strings is "a".
Input 2:
A = ["abab", "ab", "abcd"];
Output 2:
"ab"
Explanation 2:
Longest common prefix of all the strings is "ab".
Method
Find the shortest string, and then check character one by one.
Solution:
Time: O(n * length of min string) Space: O(length of min string)
public class Solution {
public String longestCommonPrefix(ArrayList<String> A) {
int len = Integer.MAX_VALUE;
for (String s : A) {
if (s.length() < len) {
len = s.length();
}
}
StringBuilder prefix = new StringBuilder();
for (int i = 0; i < len; i ++) {
char c = A.get(0).charAt(i);
for (int j = 1; j < A.size(); j ++) {
char n = A.get(j).charAt(i);
if (c != n) {
return prefix.toString();
}
}
prefix.append(c);
}
return prefix.toString();
}
}
class Solution {
public String longestCommonPrefix(String[] strs) {
if (strs == null || strs.length == 0) return "";
StringBuilder sb = new StringBuilder();
int j = 0;
String first = strs[0];
while (j < first.length()) {
char c = first.charAt(j);
boolean exit = false;
for (int i = 1; i < strs.length; i ++) {
if (j >= strs[i].length() || strs[i].charAt(j) != c) {
exit = true;
break;
}
}
if (exit) break;
sb.append(c);
j ++;
}
return sb.toString();
}
}