LeetCode-Substring with Concatenation of All Words

作者：disappearedgod

文章出處：http://blog.csdn.net/disappearedgod/article/details/39780829

時間：2014-10-4

題目

Substring with Concatenation of All Words

 Total Accepted: 13715 Total Submissions: 76035My Submissions

You are given a string, S, and a list of words, L, that are all of the same length. Find all starting indices of substring(s) in S that is a concatenation of each word in L exactly once and without any intervening characters.

For example, given:
S: "barfoothefoobarman"
L: ["foo", "bar"]

You should return the indices: [0,9].
(order does not matter).

解法

最小滑動窗口(Java AC的代碼是448ms)

因爲L中所有單詞的長度是一樣的，這樣根據wordLen，可以將S分爲wordLen組，實際意思是這樣的。
以題目中barfoothefoobarman舉例，L中單詞長度爲3，可以分爲
bar|foo|the|foo|bar|man
ba|rfo|oth|efo|oba|rma|n
b|arf|oot|hef|oob|arm|an
這樣，針對每個分組，可以利用最小滑動窗口的思想，快速的判斷是否包含需要的字符串。
直觀上來看，1和2好像都是需要從每個字符開始搜索，實際上，2利用兩個指針去在S中尋找滿足條件的字符串，並且是每次+wordLen，而且不會重複的去統計，節省了很多時間。

代碼

public class Solution {
    public List<Integer> findSubstring(String S, String[] L) {
        List<Integer> retlist = new ArrayList<Integer>();
        int wordLen = L[0].length();
        int size = L.length;
        int slen = S.length();
        int max = slen - wordLen + 1;
        if(size == 0)
            return retlist;
        int len_match = wordLen * L.length;
        if(len_match > S.length())
            return retlist;
        
        HashMap<String, Integer> map = new HashMap<String, Integer>();
        for(int i = 0; i< size; ++i){
            if(map.containsKey(L[i]))
                map.put(L[i],map.get(L[i])+1);
            else
                map.put(L[i], 1);
        }
        
        for(int i = 0; i < wordLen; i++){
            //boolean flag = true;
            HashMap<String, Integer> tmpMap = new HashMap<String, Integer>();
            int count = 0;
            int start = i;
            for(int j = start; j < max; j += wordLen){
                String str = S.substring(j, j + wordLen);
                if(!map.containsKey(str)){
                    tmpMap.clear();
                    count = 0;
                    start = j + wordLen;
                    continue;
                }
                //map has this str
                if(tmpMap.containsKey(str))
                    tmpMap.put(str, tmpMap.get(str)+1);
                else
                    tmpMap.put(str, 1);
                
                if(tmpMap.get(str) <= map.get(str)){
                    count++;
                }else{
                    while(tmpMap.get(str) > map.get(str)){
                        str = S.substring(start, start + wordLen);
                        tmpMap.put(str, tmpMap.get(str) - 1);
                        if(tmpMap.get(str) < map.get(str))
                            count--;
                        start +=wordLen;
                    }
                }
                if(count == size){
                    retlist.add(start);
                    str = S.substring(start, start + wordLen);
                    tmpMap.put(str, tmpMap.get(str)-1);
                    count--;
                    start += wordLen;
                }
                
            }
        }
        return retlist;
        
    }
}

結果

Submit Time	Status	Run Time	Language
15 minutes ago	Accepted	480 ms	java

返回

LeetCode Solution(持續更新，java>c++)

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【LeetCode】Substring with Concatenation of All Words