标签:
滑动窗口,但是很繁琐
public class Solution { public ArrayList<Integer> findSubstring(String S, String[] L) { //http://www.cnblogs.com/springfor/p/3872516.html ArrayList<Integer> res = new ArrayList<Integer>(); if(S==null || L==null || S.length()==0 || L.length==0 ||L[0].length()==0) return res; HashMap<String, Integer> dict = new HashMap<String, Integer>(); int wl = L[0].length(); for(String i: L){ if(dict.containsKey(i)){ dict.put(i, dict.get(i)+1); }else dict.put(i, 1); } for(int i=0; i<wl; i++){ int ind = i; // window‘s starting index int cnt = 0; // check how many words included in the window, and it should not be more than L‘s length HashMap<String, Integer> curdic = new HashMap<String, Integer>(); // for(int j = 0; j<S.length()-wl; j+=wl){ // check all words in S for(int j = i; j<=S.length()-wl; j+=wl){ String curwd = S.substring(j, j+wl); if(!dict.containsKey(curwd)){ // means there is no such word in L, clear curdic and move on(Concatenation must be continuous), curdic.clear(); cnt=0; ind = j+wl; }else{ if(!curdic.containsKey(curwd)){ curdic.put(curwd, 1); }else{ curdic.put(curwd,curdic.get(curwd)+1); } // is the window curdic overflow? now check cnt if(curdic.get(curwd)<=dict.get(curwd)){ cnt++; }else{ while(curdic.get(curwd)>dict.get(curwd)){ // slide the window and remove the words(tmp) till curwd String tmp = S.substring(ind, ind+wl); curdic.put(tmp, curdic.get(tmp)-1); ind += wl; if(curdic.get(tmp)<dict.get(tmp)) cnt--; } } // is cnt already == L? which means the window has all L‘s words if(cnt==L.length){ res.add(ind); String tmp = S.substring(ind, ind+wl); curdic.put(tmp, curdic.get(tmp)-1); ind += wl; cnt--; // 注意这里cnt的挪动,因为同一个i起始点,window curdic可以挪,cnt是决定是否记录一下这个位置 } } } } return res; } }
天题系列:Substring with Concatenation of All Words
标签:
原文地址:http://www.cnblogs.com/jiajiaxingxing/p/4443280.html