[LeetCode294] Flip Game II_leetcode 294 filp game ii python backtrack-优快云博客

本文探讨了翻转游戏中如何通过策略确保胜利，并提供了算法实现。通过改变连续的两个 '+' 符号为 '--'，直到无法进行操作，确定初始玩家是否能确保获胜。

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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 twoconsecutive "++" 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.

solution:

recursive and backtracking, after you move the game, it's left a non-movable result, you will win. Also, here use array to make change and restore. Look at code, there is one line

boolean win = !canWin(cs);

For example"+++", after first move, --+ is non-movable

public void test() {
        //win ++, +++, ++++, + ++ ++ +++
        //failed,+++++,++--++--
        String[] testcase = {"+","-","++","--","+++","++++","+--+","-+-+","+++++"};
        for(int i=0;i<testcase.length;i++){
            System.out.println(firstPlayWinning(testcase[i]));
        }
    }
    public boolean firstPlayWinning(String s1) {
        return canWin(s1.toCharArray());
    }
    public boolean canWin(char[] cs) {
        for(int i=0;i<cs.length-1;i++) {
            if(cs[i] == cs[i+1] && cs[i]=='+') {
                cs[i] = '-';
                cs[i+1] = '-';
                boolean win = !canWin(cs);
                cs[i] = '+';
                cs[i+1] = '+';
                if(win) return true;
            }
        }
        return false;
    }