ou are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.
For example, given:
s: "barfoothefoobarman"
words: ["foo", "bar"]
You should return the indices: [0,9]
.
(order does not matter).
所有的循环只为寻找答案, 所有的判断只为选择正确答案
public List<Integer> findSubstring(String s, String[] words) { List<Integer> res = new LinkedList<Integer>(); if(words == null || words.length == 0 || s == null || s.equals("")) return res; HashMap<String, Integer> freq = new HashMap<String, Integer>(); // 统计数组中每个词出现的次数,放入哈希表中待用 for(String word : words){ freq.put(word, freq.containsKey(word) ? freq.get(word) + 1 : 1); } // 得到每个词的长度 int len = words[0].length(); // 错开位来统计 for(int i = 0; i < len; i++){ // 因为要一单词长度递增, 因此i < len
// 建一个新的哈希表,记录本轮搜索中窗口内单词出现次数 HashMap<String, Integer> currFreq = new HashMap<String, Integer>(); // start是窗口的开始,count表明窗口内有多少词 int start = i, count = 0; for(int j = i; j <= s.length() - len; j += len){ String sub = s.substring(j, j + len); // 看下一个词是否是给定数组中的 if(freq.containsKey(sub)){ // 窗口中单词出现次数加1 currFreq.put(sub, currFreq.containsKey(sub) ? currFreq.get(sub) + 1 : 1); count++; // 如果该单词出现次数已经超过给定数组中的次数了,说明多来了一个该单词,所以要把窗口中该单词上次出现的位置及之前所有单词给去掉 while(currFreq.get(sub) > freq.get(sub)){ String leftMost = s.substring(start, start + len); currFreq.put(leftMost, currFreq.get(leftMost) - 1); start = start + len; count--; } // 如果窗口内单词数和总单词数一样,则找到结果 if(count == words.length){ String leftMost = s.substring(start, start + len); currFreq.put(leftMost, currFreq.get(leftMost) - 1); res.add(start); start = start + len; count--; } // 如果截出来的单词都不在数组中,前功尽弃,重新开始 } else { currFreq.clear(); start = j + len; count = 0; } } } return res; }
复杂的窗口指针, 而非简单的dfs, 如何设计状态来将符合题意的选项选出来, 并且如何根据指针的移动更新状态, 此处指针的设计颇为tricky, 是start 和j, 而i 是三组j而已
没要求顺序, 因此用单个计数和总和来判断是否满足题意, 关键是题意的转换