wordpress网站关闭谷歌收录汕头网站排名优化
题目：
给定一个字符串
s和一些长度相同的单词words。找出 s 中恰好可以由 words 中所有单词串联形成的子串的起始位置。
注意子串要与 words 中的单词完全匹配，中间不能有其他字符，但不需要考虑 words 中单词串联的顺序。
示例 1：
输入：
s = “barfoothefoobarman”,
words = [“foo”,“bar”]
输出：[0,9]
解释：
从索引 0 和 9 开始的子串分别是 “barfoo” 和 “foobar” 。
输出的顺序不重要, [9,0] 也是有效答案。示例 2：
输入：
s = “wordgoodgoodgoodbestword”,
words = [“word”,“good”,“best”,“word”]
输出：[]
思路
简单的动态规划。
后面再补充讲解。
代码：
public class Q0030 {public static void main(String[] args) {demo1();demo2();demo3();}private static void demo1() {String s = "wordgoodgoodgoodbestword";String[] words = {"word", "good", "best", "word"};List<Integer> substring = findSubstring(s, words);System.out.println(substring);}private static void demo2() {String s = "barfoothefoobarman";String[] words = {"foo", "bar"};List<Integer> substring = findSubstring(s, words);System.out.println(substring);}private static void demo3() {String s = "wordgoodgoodgoodbestword";String[] words = {"word", "good", "best", "good"};List<Integer> substring = findSubstring(s, words);System.out.println(substring);}public static List<Integer> findSubstring(String s, String[] words) {List<Integer> result = new ArrayList<Integer>();int length = words[1].length();// i 起始位置for (int i = 0; i < s.length() - length; i++) {List<String> wordsList = new ArrayList<>();for (String word : words) {wordsList.add(word);}StringBuffer stringBuffer = new StringBuffer();for (int after = 0; after < length; after++) {char x = s.charAt(i + after);stringBuffer.append(x);}String string = stringBuffer.toString();if (!wordsList.contains(string)) {continue;} else {int flag = 1;if (flag == 0) {continue;}for (int j = i; j < s.length() - length; j += length) {StringBuffer stringBuffer1 = new StringBuffer();for (int after = 0; after < length; after++) {char x = s.charAt(i + after);stringBuffer1.append(x);}String string1 = stringBuffer1.toString();if (wordsList.contains(string1)) {wordsList.remove(string1);if (wordsList.isEmpty()) {flag = 0;result.add(i);}continue;} else {flag = 0;break;}}}}return result;}
}
Over~