思路就是:
谁面对4块石头的时候,谁就输
(因为每次就是1-3块石头,如果剩下4块石头,你怎么拿,我都能把剩下的拿走,所以你就要想尽办法让对面面对4块石头的倍数,
但是假如你面对了4的倍数,你铁定输,因为对方也是聪明人。
于是先手能不能赢,就看
class Solution {
public boolean canWinNim(int n) {
return n % 4 != 0 ;
}
}
int nums = [ 1, 5, 8, 6 ]
我和你进行捡石头游戏,假如有4堆石头,
第一堆有1个石头,
第二堆有5个石头,
第三堆有8个石头,
第四堆有6个石头,
每次只能从最前面或者最后面取1堆石头,能否保证先手一定能赢?
分析如下:
public static void main(String[] args) {
int[] nums = {1, 5, 8, 6};
int[] nums2 = {3, 9, 1, 2};
int[] nums3 = {1, 1, 1, 1};
int[] nums4 = {2, 5, 1, 3, 7, 8, 9, 11};
int[] nums5 = {1000,0,10000,2,1};
int[] nums6 = {10, 8, 20, 15, 3};
int[] nums7 = {1, 1, 1, 10};
// int[] nums0 = {5, 8, 6};
// System.out.println(firstHandCanScore(nums0));
System.out.println(firstHandCanScore(nums));
System.out.println(firstHandCanScore(nums2));
System.out.println(firstHandCanScore(nums3));
System.out.println(firstHandCanScore(nums4));
System.out.println(firstHandCanScore(nums5));
}
private static boolean firstHandCanScore(int[] nums) {
WinScoreData winScoreData = process(nums, 0, nums.length - 1);
System.out.println(winScoreData.winScore);
return winScoreData.winScore > 0;
}
private static WinScoreData process(int[] nums, int fromIndex, int toIndex) {
if (fromIndex == toIndex) {
return new WinScoreData(nums, fromIndex, toIndex, nums[fromIndex]);
}
int startLeft = nums[fromIndex];
WinScoreData chooseLeftWinScore = process(nums, fromIndex + 1, toIndex);
int leftWinScore = startLeft - chooseLeftWinScore.winScore; // 选左边之后的赢面
int startRight = nums[toIndex];
WinScoreData chooseRightWinScore = process(nums, fromIndex, toIndex - 1);
int rightWinScore = startRight - chooseRightWinScore.winScore; // 选右边之后的赢面
int winScore = Math.max(leftWinScore, rightWinScore);
return new WinScoreData(nums, fromIndex, toIndex, winScore);
}
@AllArgsConstructor
public static class WinScoreData {
private int[] nums;
private int fromIndex;
private int toIndex;
private int winScore;
}