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 }