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30. Substring with Concatenation of All Words
一、题目
1、描述
2、题意
字符串数组元素随意顺序全部拼接,求 s 中包含拼接后字符串的所有开始索引
二、解答
1、思路:
先将数组中元素进行全排序进行拼接,再遍历字符串 s 中所有包含的开始下标。
// 30. Substring with Concatenation of All Words public static List<Integer> findSubstring(String s, String[] words) { if(s.equals("")) return null; List<Integer> resultList = new ArrayList<Integer>(); List<String> allStrlist = new ArrayList<String>(); getAllStrList(allStrlist, words, 0 ); int len = words[0].length() * words.length; for(int i = 0; i <= s.length() - len; i++) { for(String tmp: allStrlist) { if(s.subSequence(i,i+len).equals(tmp)) { resultList.add(i); break; } } } for(int i : resultList) System.out.println(i); return resultList; } //DFS: 全排列 private static void getAllStrList(List<String> allStrlist, String[] words, int index) { if(index == words.length) { StringBuilder sb = new StringBuilder(); for(String s: words) sb.append(s); allStrlist.add(sb.toString()); } for (int i = index; i < words.length; i++) { swap(words, i, index); getAllStrList(allStrlist, words, index+1); swap(words, i, index); } } private static void swap(String[] words, int i, int index) { String temp = words[i]; words[i] = words[index]; words[index] = temp; }
方法二:
将字符串数组放入一个Map 中,遍历 字符串 s 来进行判断是否为满足的情况,用到了一个临时Map 来进行记录
public static List<Integer> findSubstring2(String s, String[] words) { List<Integer> list = new ArrayList<Integer>(); Map<String, Integer> map = new HashMap<String, Integer>(); Map<String, Integer> tmp = new HashMap<String, Integer>(); int sLength = s.length(); int wordsNum = words.length; int wordslength = words[0].length(); int j; if(sLength < wordsNum || wordsNum == 0) return list; for(int i = 0; i < wordsNum; i++) { if(map.containsKey(words[i])) map.put(words[i], map.get(words[i]) + 1); else map.put(words[i], 1); } for (int i = 0; i <= sLength-wordsNum*wordslength; i++) { tmp.clear(); for(j = 0; j < wordsNum; j++) { String word = s.substring(i+j*wordslength, i + j*wordslength+wordslength); if(!map.containsKey(word)) break; // 可能有重复 if(tmp.containsKey(word)) tmp.put(word, tmp.get(word)+1); else tmp.put(word, 1); if(tmp.get(word) > map.get(word)) break; } if(j == wordsNum) list.add(i); } return list; }
30. Substring with Concatenation of All Words
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