leetcode 102: Substring with Concatenation of All Words

Substring with Concatenation of All WordsFeb 24 '12

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).

[思路]

brute force, 按顺序一个一个查过去. 因为L是无序且有可能重复, 所以建一个map记录L中word的个数. 没遇到一个word删除, 当map为空时, 即找到一个substring满足要求.

public class Solution {    public ArrayList<Integer> findSubstring(String S, String[] L) {        // Start typing your Java solution below        // DO NOT write main() function        ArrayList<Integer> res = new ArrayList<Integer>();        if(S==null || L==null || L.length==0) return res;                HashMap<String, Integer> map = new HashMap<String, Integer>();                for(int i=0; i<L.length; i++){            if(map.containsKey( L[i] ) ){                map.put(L[i], map.get(L[i])+1);            } else {                map.put(L[i], 1);            }        }                int sz= S.length();        int m=L[0].length();        int n=L.length;                for(int i=0; i<sz-n*m+1;i++) {            HashMap<String, Integer> temp = new HashMap<String, Integer>(map);            for(int j=i; j<i+n*m;j+=m) {                String t = S.substring(j,j+m);                if( temp.containsKey(t) ){                    int x=temp.get(t);                    if(x==1) temp.remove(t);                    else temp.put(t, x-1);                } else {                    break;                }                             if(temp.isEmpty()) {                    res.add(i);                }            }        }        return res;    }}