leetCode 30.Substring with Concatenation of All Words (words中全部子串相连) 解题思路和方法

Substring with Concatenation of All Words 

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 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).

思路：leetcode 上有些题通过率低，并不见得是算法难，我认为非常大一部分原因是题目描写叙述不清晰。导致规则理解不透彻，所以编程的时候就会发生规则理解偏差的问题。

本题也是一样，刚一開始对规则理解偏差比較多，以为wors中字符出如今子串中出现一次就可以，无论反复还是不反复，可是后面提交只是时，看case全然理解错了规则，仅仅能又一次改写代码，非常麻烦。

怨言非常大，规则制定和说明也是非常重要的一点。

代码例如以下：

public class Solution {
    public List<Integer> findSubstring(String s, String[] words) {
        List<Integer> list = new ArrayList<Integer>();
        if(words.length == 0 || s.length() == 0){
            return list;
        }
        Map<String,Integer> map = new HashMap<String,Integer>();//保存个数以及值
        for(int i = 0; i < words.length; i++){
            if(map.get(words[i]) == null){
            	map.put(words[i],1);//将word保存
            }else{
               map.put(words[i],map.get(words[i])+1);//将word保存的数值+1 
            }
        }
        Map<String,Integer> mapValue = new HashMap<String,Integer>(map);//保存反复的个数,方便又一次赋值
        int len = words.length;//去除反复之后的数组长度
        
        int wordLen = words[0].length();
        String temp = "";
        int count = 0;//每一个单词出现一次，就记录一次
        for(int i = 0; i <= s.length() - len*wordLen;i++){
        	count = 0;//初始化
            for(int j = 0; j < len;j++){
                temp = s.substring(i + j * wordLen,i + (j+1) * wordLen);//截取wordLen长的字符串
                if(map.get(temp) != null && map.get(temp) != 0){//假设map还有多于0个
                    map.put(temp,map.get(temp)-1);//map中数值减去1
                    count++;//记录数+1
                }else{
                    break;
                }
            }
            if(count == len){//假设记录数与len相等。则说明符合要求
                list.add(i);
            }
            //HashMap又一次初始化
            for(String key:map.keySet()){//这样更高速
                map.put(key,mapValue.get(key));
            }
        }
        return list;
    }
}