[LeetCode][Java] Substring with Concatenation of All Words
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题目:
You 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 wordsexactly 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).
题意:
给定一个字符串s, 和一组字符串数组words,数组中的所有字符串长度都一样。从字符串s中找出所有子串的开始标号,这些子串是words中所有字符串的组合,并且每个字符串只出现一次且没有其他字符插入他们之间,这些字符串的排列顺序无所谓。
算法分析:
* 因为L中所有单词的长度是一样的,这样根据wordLen,可以将S分为wordLen组,实际意思是这样的。
* 以题目中barfoothefoobarman举例,L中单词长度为3,可以分为
* bar|foo|the|foo|bar|man
* ba|rfo|oth|efo|oba|rma|n
* b|arf|oot|hef|oob|arm|an
* 这样,针对每个分组,可以利用最小滑动窗口的思想,快速的判断是否包含需要的字符串。
* 直观上来看,是需要从每个字符开始搜索,实际上,利用两个指针去在S中寻找满足条件的字符串,并且是每次+wordLen,而且不会重复的去统 * 计,节省
* 了很多时间。
方法二:
思路仍然是维护一个窗口,如果当前单词在字典中,则继续移动窗口右端,否则窗口左端可以跳到字符串下一个单词了。假设源字符串的长度为n,字典中单词的长度为l。因为不是一个字符,所以我们需要对源字符串所有长度为l的子串进行判断。每次按顺序在源字符串中截取和字典中所有字符串长度相等的长度,判断新截取的子串和字典中字符串是否匹配,匹配就加入到结果中,不匹配就依次继续在源字符串中截取新的子串,重复上述过程直到结束。
AC代码:
<span style="font-size:12px;">public class Solution{ public ArrayList<Integer> findSubstring(String S, String[] L) { ArrayList<Integer> list = new ArrayList<Integer>();int len = L.length;if (len == 0) return list;int wordLen = L[0].length();Map<String, Integer> wordsMap = new HashMap<String, Integer>();for (int i = 0; i < len; i++) {int num = 1;if (wordsMap.get(L[i]) != null) num += wordsMap.get(L[i]);wordsMap.put(L[i], num);}int slen = S.length();int max = slen - wordLen + 1;for (int i = 0; i < wordLen; i++){Map<String, Integer> numMap = new HashMap<String, Integer>();int count = 0;int start = i;for (int end = start; end < max; end += wordLen) {String tempStr = S.substring(end, end + wordLen);if (!wordsMap.containsKey(tempStr))//给定字符串数组中不包含当前的字符串,直接跳到下一个字符串{numMap.clear();count = 0;start = end + wordLen;continue;}int num = 1;if (numMap.containsKey(tempStr)) num += numMap.get(tempStr);numMap.put(tempStr, num);if (num <= wordsMap.get(tempStr)) count++;//只有在小于给定数组元素个数的情况下才自加else {while (numMap.get(tempStr) > wordsMap.get(tempStr)) {tempStr = S.substring(start, start + wordLen);//在现在的map尾部中出现大于给定数组元素个数的情况是时,去除map头部元素numMap.put(tempStr, numMap.get(tempStr) - 1);if (numMap.get(tempStr) < wordsMap.get(tempStr)) count--;//去除了元素了,个数自来就少了一个start += wordLen;//对应的起始元素也往后移动了一个}}if (count == len) {list.add(start);tempStr = S.substring(start, start + wordLen);//满足条件后去除头个元素,也就是重新后移一个位置,看看后面的满足条件不numMap.put(tempStr, numMap.get(tempStr) - 1);count--;start += wordLen;} }}return list; }}</span>
方法二:
public class Solution { public List<Integer> findSubstring(String S, String[] L) { List<Integer> result=new ArrayList<Integer>(); if(L.length==0||S.length()==0) return result; int wordlen=L[0].length(); //map中存放L HashMap<String,Integer> map=new HashMap<String,Integer>(); for(int i=0;i<L.length;i++) { Integer value=map.get(L[i]); if(value==null) value=1; else value+=1; map.put(L[i],value); } for(int i=0;i+wordlen<=S.length();i++) { if(i + wordlen * L.length > S.length()) { break; } if(map.containsKey(S.substring(i,i+wordlen))) { boolean b=checkString(S.substring(i,i+wordlen*L.length),new HashMap<String,Integer>(map),wordlen); if(b==true) result.add(i); } } return result; } //检查字符串S是不是map中字符串的组合 public boolean checkString(String s,HashMap<String,Integer> map,int wordlen) { boolean flag=true; int i=0; while(s.length()>0) { String temp=s.substring(0,wordlen); Integer value=map.get(temp); if(value==null||value==0) { flag=false; break; }else{ value-=1; map.put(temp,value); s=s.substring(wordlen);//该子字符串从指定索引处的字符开始,直到此字符串末尾。 } } return flag; }}
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