leetcode 466. Count The Repetitions

leetcode 466. Count The Repetitions

这一题主要考察复杂问题的简化求解思路，从简单的例子着手。

public class Solution {    public int getMaxRepetitions(String s1, int n1, String s2, int n2) {        if (!ableToObtain(s1, s2)) return 0; // check if [s1. ∞] obtains s2        int cnt=0, k=-1;        String s=s1;        StringBuilder remainBuilder; // record `remain string`        ArrayList<String> stringList=new ArrayList<>(); // record all the `remain string`        ArrayList<Integer> countList=new ArrayList<>(); // record matching count from start to the current remain string         stringList.add(""); // record empty string        countList.add(0);        for (int i=0;i<=n1;i++) {            remainBuilder=new StringBuilder();            cnt+=getRemain(s, s2, remainBuilder); // get the next remain string, returns the count of matching            String remain=remainBuilder.toString();            if ((k=stringList.indexOf(remain))!=-1) break; // if there is a loop, break            stringList.add(remain); // record the remain string into arraylist             countList.add(cnt);            s=remain+s1; // append s1 to make a new string        }        // here, k is the beginning of the loop        if (k==-1) return cnt/n2; // if there is no loop        int countOfLoop=cnt-countList.get(k), loopLength=stringList.size()-k; // get matching count in the loop, and loop length        cnt=countList.get(k);        n1-=k;        cnt+=countOfLoop*(n1/loopLength);        n1%=loopLength;         cnt+=countList.get(k+n1)-countList.get(k);        return cnt/n2;    }    // check if [s1. ∞] obtains s2    private boolean ableToObtain(String s1, String s2) {        boolean[] cnt=new boolean[26];        for (char c: s1.toCharArray()) cnt[c-'a']=true;        for (char c: s2.toCharArray()) {            if (!cnt[c-'a']) return false;        }        return true;    }    // get remain string after s1 obtains s2, return the matching count    private int getRemain(String s1, String s2, StringBuilder remain) {        int cnt=0, lastMatch=-1, p2=0;        for (int p1=0;p1<s1.length();p1++) {            if (s1.charAt(p1)==s2.charAt(p2)) {                if (++p2==s2.length()) {                    p2=0;                    cnt++;                    lastMatch=p1;                }            }        }        remain.append(s1.substring(lastMatch+1));        return cnt;    }}