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;
}
}
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