Leetcode 14 Longest Common Prefix

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

方法一： $L C P (S 1 \dots S n ) = L C P (L C P (L C P (S 1 , S 2 ), S 3 ), \dots S n )$

$第一种$

public class Solution {    public String longestCommonPrefix(String[] strs) {        if (strs == null)            return null;        else if(strs.length == 0){            return "";        }        else{             StringBuilder str = new StringBuilder(strs[0]);             for (int i = 1; i < strs.length; i++){             str = new StringBuilder(prefix(str.toString(),strs[i]));             }          return str.toString();            }    }    public String prefix(String str1, String str2){       StringBuilder str = new StringBuilder();       for (int i = 0; i < str1.length() && i < str2.length(); i++){           if(str1.charAt(i) == str2.charAt(i))               str.append(str1.charAt(i));           else             break;       }           return str.toString();     }}

第二种

public class Solution{public String longestCommonPrefix(String[] strs) {    if (strs.length == 0) return "";    String prefix = strs[0];    for (int i = 1; i < strs.length; i++)        while (strs[i].indexOf(prefix) != 0) {            prefix = prefix.substring(0, prefix.length() - 1);            if (prefix.isEmpty()) return "";        }            return prefix; }}

方法二：分治法

public class Solution{public String longestCommonPrefix(String[] strs) {    if (strs == null || strs.length == 0) return "";            return longestCommonPrefix(strs, 0 , strs.length - 1);}private String longestCommonPrefix(String[] strs, int l, int r) {    if (l == r) {        return strs[l];    }    else {        int mid = (l + r)/2;        String lcpLeft =   longestCommonPrefix(strs, l , mid);        String lcpRight =  longestCommonPrefix(strs, mid + 1,r);        return commonPrefix(lcpLeft, lcpRight);   }}String commonPrefix(String left,String right) {    int min = Math.min(left.length(), right.length());           for (int i = 0; i < min; i++) {        if ( left.charAt(i) != right.charAt(i) )            return left.substring(0, i);    }    return left.substring(0, min); }}

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