Longest common prefix --leetCode

package com.helloxin.leetcode.algorithms;

/**
 *  create by one leaf on 2017/12/12
 */
public class Longest_common_prefix {

    /**
     * 根據String的indexOf和substring 來找他們的公衆部分
     * @param strs
     * @return
     */
    public static String longestCommonPrefix(String[] strs) {
        if (strs.length == 0) {
            return "";
        }
        String prefix = strs[0];
        for (int i = 1; i < strs.length; i++) {
            //沒有子集會返回 -1，0是第一個位置 startWith
            while (strs[i].indexOf(prefix) != 0) {
                prefix = prefix.substring(0, prefix.length() - 1);
                //若是爲空 那麼沒有公衆前綴
                if (prefix.isEmpty()) {
                    return "";
                }
            }
        }
        return prefix;
    }

    /**
     * 這個方法寫的就比較直觀了 取第一次字符，去遍歷其餘的字符串  看是否符合
     * @param strs
     * @return
     */
    public static String longestCommonPrefix2(String[] strs) {
        if (strs == null || strs.length == 0) {
            return "";
        }
        //遍歷 第一個字符串
        for (int i = 0; i < strs[0].length() ; i++){
            char c = strs[0].charAt(i);
            //開始比對
            for (int j = 1; j < strs.length; j ++) {
                //若是到了某個字符串的長度 或者已經不匹配
                if (i == strs[j].length() || strs[j].charAt(i) != c) {
                    return strs[0].substring(0, i);
                }
            }
        }
        return strs[0];
    }

    /**
     * Divide and conquer  這是看solution 裏面 前輩們寫的第三種方法  想不到 並且我也不習慣使用遞歸的編程
     * 或許是由於遞歸 沒有那麼直觀看到怎麼執行的
     * 用於這個道題目到沒看出什麼優點 可是熟悉這個概念還不錯
     *
     * @param strs
     * @return
     */
    public static String longestCommonPrefix3(String[] strs) {
        if (strs == null || strs.length == 0) {
            return "";
        }
        return longestCommonPrefix3(strs, 0 , strs.length - 1);
    }

    private static String longestCommonPrefix3(String[] strs, int l, int r) {
        if (l == r) {
            return strs[l];
        }
        else {
            int mid = (l + r)/2;
            //遞歸調用
            String lcpLeft =   longestCommonPrefix3(strs, l , mid);
            String lcpRight =  longestCommonPrefix3(strs, mid + 1,r);
            return commonPrefix(lcpLeft, lcpRight);
        }
    }

    static 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);
    }

    /**
     * Binary search  在什麼狀況下  須要考慮這麼寫呢 這是solution中提供的第四種方法
     * @param strs
     * @return
     */
    public static String longestCommonPrefix4(String[] strs) {
        if (strs == null || strs.length == 0) {
            return "";
        }
        int minLen = Integer.MAX_VALUE;
        for (String str : strs) {
            minLen = Math.min(minLen, str.length());
        }
        int low = 1;
        int high = minLen;
        while (low <= high) {
            int middle = (low + high) / 2;
            if (isCommonPrefix(strs, middle)) {
                low = middle + 1;
            } else {
                high = middle - 1;
            }
        }
        return strs[0].substring(0, (low + high) / 2);
    }

    //判斷是否有共同前綴
    private static boolean isCommonPrefix(String[] strs, int len){
        String str1 = strs[0].substring(0,len);
        for (int i = 1; i < strs.length; i++) {
            if (!strs[i].startsWith(str1)) {
                return false;
            }
        }
        return true;
    }




    public static void main(String[] args) {
        System.out.println(longestCommonPrefix3(new String[]{"YE","YE"}));
    }
}

https://github.com/woshiyexinjie/leetcode-xin