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LintCode Coins in a Line II

news 来源：原创 2023/3/28 5:02:31

原题链接在这里：http://www.lintcode.com/en/problem/coins-in-a-line-ii/

题目：

There are n coins with different value in a line. Two players take turns to take one or two coins from left side until there are no more coins left. The player who take the coins with the most value wins.

Could you please decide the first player will win or lose?

Example

Given values array A = [1,2,2], return true.

Given A = [1,2,4], return false.

题解：

是Coins in a Line的进阶题.

还是DP. 状态dp[i]是还剩i枚硬币，先手的人最后能拿到最多的值.

转移方程 dp[i] = max(min(dp[i-2],dp[i-3])+values[n-i], min(dp[i-3],dp[i-4])+values[n-i]+values[n-i+1])

初始化dp[0] = 0, dp[1] = values[i-1], dp[2] = values[i-1]+values[i-2], dp[3] = values[i-2] + values[i-3].

答案 dp[n] > sum/2

Time Complexity: O(n). Space: O(n), stack space.

AC Java:

 1 public class Solution {
 2     public boolean firstWillWin(int[] values) {
 3         if(values == null || values.length == 0){
 4             return false;
 5         }
 6         
 7         int n = values.length;
 8         int [] dp = new int[n+1];
 9         boolean [] used = new boolean [n+1];
10         int sum = 0;
11         for(int value : values){
12             sum += value;
13         }
14         
15         return sum < 2*memorySearch(values, dp, n, used);
16     }
17     
18     private int memorySearch(int [] values, int [] dp, int n, boolean [] used){
19         if(used[n]){
20             return dp[n];
21         }
22         used[n] = true;
23         
24         if(n<=0){
25             dp[n] = 0;;
26         }else if(n == 1){
27             dp[n] = values[values.length-1];
28         }else if(n == 2){
29             dp[n] = values[values.length-1] + values[values.length-2];
30         }else if(n == 3){
31             dp[n] = values[values.length-2] + values[values.length-3];
32         }else{
33             dp[n] = Math.max(
34                 Math.min(memorySearch(values, dp, n-2, used), memorySearch(values, dp, n-3, used)) + values[values.length-n]
35                 , Math.min(memorySearch(values, dp, n-3, used), memorySearch(values, dp, n-4, used)) + values[values.length-n] + values[values.length-n+1]
36                 );
37         }
38         return dp[n];
39     }
40 }