Longest Common Prefix

Write a function to find the longest common prefix string amongst an array of strings.
Solution:
就是brute force的想法，以第一个字符串为标准，对于每个字符串从第一个字符开始，看看是不是和标准一致，如果不同，则跳出循环返回当前结果，否则继续下一个字符。
时间复杂度应该是O(m*n)，m表示字符串的最大长度，n表示字符串的个数，空间复杂度应该是O(m),即字符串的长度.

public class Solution {
    public String longestCommonPrefix(String[] strs) {
        if(strs == null || strs.length == 0 || hasEmpty(strs))  //if strs has empty string, return ""
            return "";
        if(strs.length == 1) 
            return strs[0];
        
        String str1 = strs[0];
        for (int i = 0; i < str1.length(); i++) {
            char temp = str1.charAt(i); 
            
            for (int j = 1; j < strs.length; j++) {
                if (i > strs[j].length() - 1 || strs[j].charAt(i) != temp) {
                    return str1.substring(0, i);
                }
            }
        }
        return str1;
    }
    
    public boolean hasEmpty(String[] strs){
        for(int i = 0; i < strs.length; i++){
            if(strs[i].length() == 0)
                return true;
        }
        return false;
    }
}

Or

public class Solution {
    public String longestCommonPrefix(String[] strs) {
        StringBuilder res = new StringBuilder();
        if(strs == null || strs.length == 0)
            return res.toString();
        
        int index = 0;
        boolean isSame = true;
        while(isSame) {
            for(int i = 0; i < strs.length; i++) {
                if(strs[i].length() <= index || strs[i].charAt(index) != strs[0].charAt(index)) {
                    isSame = false;
                    break;
                }
            }
            if(isSame) {
                res.append(strs[0].charAt(index));
                index++;
            }
        }
        return res.toString();
    }
}