package com.example.demo;public class Test14 {/*** 查找字符串数组中最长的公共**前缀**** @param strs* @return*/public String longestCommonPrefix(String[] strs) {//水平扫描法if (strs == null || strs.length == 0) {return "";}String prefix = strs[0];for (int i = 1; i < strs.length; i++) {// 因为查找的各个字符串的前缀，所以A字符串在B字符串中的开头，不是开头的话，prefix--，重新查询，制动为空时结束while (strs[i].indexOf(prefix) != 0) {prefix = prefix.substring(0, prefix.length() - 1);if (prefix.isEmpty()) {return "";}}}return prefix;}public static void main(String[] args) {Test14 t = new Test14();String[] strings = {"abcd", "abc", "ab"};String s = t.longestCommonPrefix(strings);System.out.println(s);}
}

package com.example.demo;public class Test14 {/*** 查找字符串数组中最长的公共**前缀**** @return*/public String longestCommonPrefix1(String[] strs) {//分治法if (strs == null || strs.length == 0) {return "";}return longestCommonPrefix1(strs, 0, strs.length - 1);}/*** 递归、分治** @param strs* @param left* @param right* @return*/private String longestCommonPrefix1(String[] strs, int left, int right) {// 递归终止条件if (left == right) {// 递归的终止条件要选择好return strs[left];} else {int mid = (left + right) / 2;// 递归的思想（不要深入想递归内部，只要在宏观上认为，该递归返回的字符串就是left到mid之间的最长公共前缀）String lcpLeft = longestCommonPrefix1(strs, left, mid);String lcpRight = longestCommonPrefix1(strs, mid + 1, right);//将返回的左右两个最长公共子串进行比较，获取最终的最长公共前缀return comparePreFix(lcpLeft, lcpRight);}}/*** 比较两个字符串中的最长前缀** @param lcpLeft* @param lcpRight* @return*/private String comparePreFix(String lcpLeft, String lcpRight) {int len = Math.min(lcpLeft.length(), lcpRight.length());for (int i = 0; i < len; i++) {if (lcpLeft.charAt(i) != lcpRight.charAt(i)) {return lcpLeft.substring(0, i);}}return lcpLeft.substring(0, len);}public static void main(String[] args) {Test14 t = new Test14();String[] strings = {"abcd", "abc", "ab"};String s = t.longestCommonPrefix1(strings);System.out.println(s);}
}