294 Flip Game II

标签：searching   already   ant   cas   tle   inpu   ase   val   time   

You are playing the following Flip Game with your friend: 
Given a string that contains only these two characters: + and -, 
you and your friend take turns to flip two consecutive "++" into "--". 
The game ends when a person can no longer make a move and therefore the other person will be the winner.

Write a function to determine if the starting player can guarantee a win.

For example, given s = "++++", return true. 
The starting player can guarantee a win by flipping the middle "++" to become "+--+".

Follow up:
Derive your algorithm‘s runtime complexity.

The idea is try to replace every "++" in the current string s to "--" and see if the opponent has the chance to win or not, if the opponent is guaranteed to lose, great, we win!

For the time complexity, here is what I thought, let‘s say the length of the input string s is n, there are at most n - 1 ways to replace "++" to "--" (imagine s is all "+++..."), once we replace one "++", there are at most (n - 2) - 1 ways to do the replacement, it‘s a little bit like solving the N-Queens problem, the time complexity is (n - 1) x (n - 3) x (n - 5) x ..., so it‘sO(n!!), double factorial.

public class Solution {
    public boolean canWin(String s) {
        if (s==null || s.length()<=1) return false;
        for (int i=0; i<s.length()-1; i++) {
            if (s.charAt(i)==‘+‘ && s.charAt(i+1)==‘+‘ && !canWin(s.substring(0,i) + "--" +     
                                                                                 s.substring(i+2))) 
                return true;
        }
        return false;
    }
}

Better Solution:  (205ms -> 19ms)

 but the time complexity of the backtracking method is high. During the process of searching, we could encounter duplicate computation as the following simple case.

One search path:

Input s = "++++++++"

Player 0: "--++++++"

Player 1: "----++++"

Player 0: "----+--+"

Player0 can win for the input string as "----++++".

Another search path:

Player 0: "++--++++"

Player 1: "----++++"

Player 0: "----+--+"

(Duplicate computation happens. We have already known anyone can win for the

input string as "----++++".)

Use a HashMap to avoid duplicate computation

Key : InputString.

Value: can win or not.

public boolean canWin(String s) {
    if (s == null || s.length() < 2) {
        return false;
    }
    HashMap<String, Boolean> winMap = new HashMap<String, Boolean>();
    return helper(s, winMap);
}

public boolean helper(String s, HashMap<String, Boolean> winMap) {
    if (winMap.containsKey(s)) {
        return winMap.get(s);
    }
    for (int i = 0; i < s.length() - 1; i++) {
        if (s.startsWith("++", i)) {
            String t = s.substring(0, i) + "--" + s.substring(i+2);
            if (!helper(t, winMap)) {
                winMap.put(s, true);
                return true;
            }
        }
    }
    winMap.put(s, false);
    return false;
}

　　只要任何一步能让下一步的对手false, 就返回ture, 不然没有任何一步, 就返回false, 像下棋一样

294 Flip Game II

标签：searching   already   ant   cas   tle   inpu   ase   val   time   

原文地址：http://www.cnblogs.com/apanda009/p/7395828.html