030 Substring with Concatenation of All Words 与所有单词相关联的字串

给定一个字符串 s 和一些长度相同的单词 words，找出 s 与 words 中所有单词（words 每个单词只出现一次）串联一起（words中组成串联串的单词的顺序随意）的字符串匹配的所有起始索引，子串要与串联串完全匹配，中间不能有其他字符。
举个例子，给定：
s："barfoothefoobarman"
words：["foo", "bar"]
你应该返回的索引: [0,9]。（任意顺序）
详见：https://leetcode.com/problems/substring-with-concatenation-of-all-words/description/

Java实现：

class Solution {public List<Integer> findSubstring(String s, String[] words) {List<Integer> res=new ArrayList<Integer>();if(s.isEmpty()||s==null||words==null||words.length==0){return res;}int n=words.length;int m=words[0].length();Map<String,Integer> m1=new HashMap<String,Integer>();for(String str:words){if(m1.containsKey(str)){m1.put(str,m1.get(str)+1);}else{m1.put(str,1);}}for(int i=0;i<=s.length()-n*m;++i){Map<String,Integer> m2=new HashMap<String,Integer>();int j=0;for(;j<n;++j){String t=s.substring(i+j*m,i+j*m+m);if(!m1.containsKey(t)){break;}if(m2.containsKey(t)){m2.put(t,m2.get(t)+1);}else{m2.put(t,1);}if(m2.get(t)>m1.get(t)){break;}}if(j==n){res.add(i);}}return res;}
}

参考：https://www.cnblogs.com/grandyang/p/4521224.html

https://blog.csdn.net/fly_yr/article/details/47957459

转载于:https://www.cnblogs.com/xidian2014/p/8687693.html

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