Leetcode: Substring with Concatenation of All Words

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"]

这道题想的时候颇费了一些周折,想过把L的所有concatenation组合出来,放到hash或map里,然后遍历S的时候直接看。但是这样如果L的size: Lsize过大的话,可能的组合有Lsize!种,组合数剧增,效率低下,所以不采用这种方法。

又考虑在S中从左向右一个word一个word遍历过去,一次只取L中一个word的大小,存入Hashmap,如果匹配到了L,则算找到了一种可能的concatenation;这样也有缺点,比如L为foo the kgm tim cog, S为bar foo tim cog kgm tim foo the, S遍历到第二个tim时出现不匹配,所以我的指针得跳到第一个tim的下一个,才能不漏掉所有的可能性。我要想跳到第一个tim得存下它的位置,这就要求hashmap的value为元素位置,这就导致我不好进行匹配,因为L若要存为一个hashmap,它的value我无法确定

所以参考网上有了第三种做法,先把L中的word存入一个Hashmap, 记录每个word的出现次数,然后在S中从左向右挨个遍历过去,每次取wsize * Lsize个字母分成Lsize份存入Hashmap中,每份大小为wsize, 然后比较这个Hashmap与L的Hashmap是否一致

 1 import java.util.*;

 2 

 3 public class Solution {

 4     public List<Integer> findSubstring(String S, String[] L) {

 5         ArrayList<Integer> result = new ArrayList<Integer>();

 6         if (S.length() == 0 || L.length == 0) return result;

 7         Hashtable<String, Integer> obj = new Hashtable<String, Integer>();

 8         Hashtable<String, Integer> real = new Hashtable<String, Integer>();

 9         int Lsize = L.length;

10         int wsize = L[0].length();

11         for (String a : L) {

12             if (obj.containsKey(a)) {

13                 int k = obj.get(a);

14                 obj.put(a, k+1);

15             }

16             else obj.put(a, 1);

17         }

18         

19         for (int i = 0; i <= S.length() - Lsize * wsize; i++) {

20             real.clear();

21             for (int j = 0; j <= Lsize - 1; j++) {

22                 String temp = S.substring(i + j * wsize, i + (j+1) * wsize);
           if (!obj.containsKey(temp)) break;
23 if (real.containsKey(temp)) { 24 int m = real.get(temp); 25 real.put(temp, m+1); 26 } 27 else real.put(temp, 1); 28 } 29 if (real.equals(obj)) { 30 result.add(i); 31 } 32 } 33 34 return result; 35 } 36 }

 另一个版本,也许更好懂一点

 1 public class Solution {

 2     public List<Integer> findSubstring(String S, String[] L) {

 3         ArrayList<Integer> res = new ArrayList<Integer>();

 4         HashMap<String, Integer> target = new HashMap<String, Integer>();

 5         HashMap<String, Integer> real = new HashMap<String, Integer>();

 6         for (String str : L) {

 7             if (target.containsKey(str)) {

 8                 int val = target.get(str);

 9                 target.put(str, val+1);

10             }

11             else {

12                 target.put(str, 1);

13             }

14         }

15         

16         int wSize = L[0].length();

17         int lSize = L.length;

18         for (int i=0; i<=S.length()-lSize*wSize; i++) {

19             for (int j=i; j<=i+lSize*wSize-1; j=j+wSize) {

20                 String temp = S.substring(j, j+wSize);

21                 if (!target.containsKey(temp)) break;

22                 if (real.containsKey(temp)) {

23                     int value = real.get(temp);

24                     real.put(temp, value+1);

25                 }

26                 else {

27                     real.put(temp, 1);

28                 }

29             }

30             if (real.equals(target)) {

31                 res.add(i);

32             }

33             real.clear();

34         }

35         return res;

36     }

37 }

 

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