剛做完美圖的筆試，兩道編程題，第一道比較簡(jiǎn)單：找出一串用逗號(hào)隔開的字符串中不重復(fù)的那個(gè)數(shù)。

以下是第二道，時(shí)間有限，很多地方?jīng)]來(lái)得及優(yōu)化，整體邏輯應(yīng)該沒(méi)錯(cuò)。

question：

You are playing the following Nim Game with your friend: There is a heap of stones on the table, each time one of you take turns to remove 1 to 3 stones. The one who removes the last stone will be the winner. You will take the first turn to remove the stones.
Both of you are very clever and have optimal strategies for the game. Write a function to determine whether you can win the game given the number of stones in the heap. True for win and false for lose.
For example, if there are 4 stones in the heap, then you will never win the game: no matter 1, 2, or 3 stones you remove, the last stone will always be removed by your friend.

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解：

public class Main {public static void main(String[] args) {Scanner scanner = new Scanner(System.in);int sum = scanner.nextInt();//一開始我拿的三種狀況if (chance(sum,1,1)) {System.out.println(true);}else if (chance(sum,2,1)) {System.out.println(true);}else if (chance(sum,3,1)) {System.out.println(true);}else {System.out.println(false);}}/*** * @param surplus 剩余的石頭數(shù)量* @param take_num 這一次拿走的數(shù)量* @param people 1代表是我，2代表是朋友* @return 返回true，則表示我能贏*/private static boolean chance(int surplus,int take_num,int people) {surplus = surplus-take_num;//當(dāng)某人拿完，剩余四塊時(shí)，該人贏if (surplus==4) {if (people==1) {return true;}else {return false;}}//當(dāng)某人拿完，剩余數(shù)小于4塊時(shí)，另一個(gè)人贏if (surplus<=3) {if (people==1) {return false;}else {return true;}}//此時(shí)剩余的石頭大于4，繼續(xù)拿if (people==1) {//交換人people = 2;//如果是朋友拿，則必須三個(gè)trueif (chance(surplus,1,people)) {if (chance(surplus,2,people)) {if (chance(surplus,3,people)) {return true;}else {return false;}}else {return false;}}else {return false;}}else {people = 1;//如果是我拿，則只需有一個(gè)true即可if (chance(surplus,1,people)) {return true;}else if (chance(surplus,2,people)) {return true;}else if (chance(surplus,3,people)) {return true;}else {return false;}}} }

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轉(zhuǎn)載于:https://www.cnblogs.com/red-code/p/6720114.html

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