leetcode -- Longest Common Prefix

Write a function to find the longest common prefix string amongst an array of strings.

方法1:BF, 首先计算两个字符串的prefix，每次用这个prefix与下一个字符串生成新的prefix

效果较差，大数据集直接挂了，代码可读性太差-_-!

public String longestCommonPrefix(String[] strs) {
        // Start typing your Java solution below
        // DO NOT write main() function
        String commonPre = null;
        int commonPreLength = 0;
        if(strs.length == 0){
            return "";
        } else if(strs.length == 1){
            return strs[0];
        } else {
            String a = strs[0];
            String b = strs[1];
            String abPrefix = findCommonPrefix(a, b);
            int abPrefixSize = abPrefix.length();
            
            int strsSize = strs.length - 2;
            for(int i = 0; i < strsSize; i++){
                String tmp = strs[i + 2];
                int tmpLength = tmp.length();
                if(tmpLength >= abPrefixSize && tmp.substring(0, abPrefixSize).equals(abPrefix))
                    continue;
                else if(tmpLength >= abPrefixSize && !tmp.substring(0, abPrefixSize).equals(abPrefix)){
                    abPrefix = abPrefix.substring(0, abPrefixSize - 1);
                    abPrefixSize = abPrefix.length();
                    i--;
                    continue;
                }    
                if(tmpLength < abPrefixSize){
                    abPrefix = abPrefix.substring(0, tmpLength);
                    abPrefixSize = abPrefix.length();
                    i--;
                    continue;
                }
                
                
                
            }
            return abPrefix;
        }
    }
    
    public String findCommonPrefix(String a, String b){
        int length = Math.min(a.length(), b.length());
        int i;
        for(i = 0; i < length; i++){
            if(a.charAt(i) != b.charAt(i)){
                break;
            }
        }
        return a.substring(0, i);
    }

 refactor code, 但BF的时间复杂度仍然有O(n^2)

public String longestCommonPrefix(String[] strs) {
        String compare = "";
        if(strs.length == 0)
            return compare;
        
        compare = strs[0];
        for(int i = 1; i < strs.length; i++){
            int j = 0;
            while(j < compare.length() && j < strs[i].length()){
                if(compare.charAt(j) != strs[i].charAt(j)){
                    break;
                }
                j++;
            }
            compare = strs[i].substring(0, j);
        }
        return compare;
    }

 方法二：依次比较每个字符串的相应位置的字符，直到遇到不匹配项。可以减少不必要的比较

   public String longestCommonPrefix(String[] strs) {
        String compare = "";
        if(strs.length == 0)
            return compare;
        
        int k = 0;
        while(true){
            if(k == strs[0].length())
                break;
            char p = strs[0].charAt(k); 
            int i;
            for(i = 1; i < strs.length; i++){
                if(k == strs[i].length())
                    break;
                
                if(p != strs[i].charAt(k)){
                    break;
                }
            }
            
            if(i != strs.length)
                break;
            
            compare += p;
            k++;
        }
        
        return compare;
    }

 refactor code

public String longestCommonPrefix(String[] strs) {
         String compare = "";
         if(strs.length == 0)
             return compare;
         
         int k = 0;
         while(true){
             char p = ' '; 
             int i;
             for(i = 0; i < strs.length; i++){
                 
                 if(k == strs[i].length())
                     break;
                     
                if(i == 0){
                    p = strs[i].charAt(k);
                }
                 
                 if(p != strs[i].charAt(k)){
                     break;
                 }
             }
             
             if(i != strs.length)
                 break;
             
             compare += p;
             k++;
         }
         
         return compare;
     }