Create Guess_Number_Game_II.java

Author: zxqiu

arrays/Guess_Number_Game_II.java

@@ -0,0 +1,73 @@
+/*
+
+Guess Number Game II
+
+
+
+We are playing the Guess Game. The game is as follows:
+I pick a number from 1 to n. You have to guess which number I picked.
+Every time you guess wrong, I'll tell you whether the number I picked is higher or lower.
+However, when you guess a particular number x, and you guess wrong, you pay $x. You win the game when you guess the number I picked.
+
+
+Example
+
+Given n = 10, I pick 8.
+First round: You guess 5, I tell you that it's higher. You pay $5.
+Second round: You guess 7, I tell you that it's higher. You pay $7.
+Third round: You guess 9, I tell you that it's lower. You pay $9.
+
+Game over. 8 is the number I picked.
+You end up paying $5 + $7 + $9 = $21.
+
+Given a particular n ≥ 1, find out how much money you need to have to guarantee a win.
+So when n = ｀10, return16`
+
+
+
+解：
+Dynanmic Programming
+
+递推公式：
+dp[i][j]表示从i到j中猜中任意数字的最小开销。
+假设猜数字k，i<k<j，那么需要在i到k-1和k+1到j中继续猜。
+依据minimax原理，对方可选择让我方开销最大的回答方式，故选择i到k-1和k+1到j中开销大的那个。
+故猜k时的最小开销是：
+k + max(dp[i][k-1], dp[k+1][j])
+
+k可以取i到j，我方应当选择让自己开销最小的一个，故：
+dp[i][j] = min(k + max(dp[i][k-1], dp[k+1][j])，i<k<j)
+
+
+初始条件：
+由于每次需要用到i和j之间长度小于当前长度的dp值，所以应当按照长度递增来进行计算。
+当i和j相等时，长度为1，猜中的开销为0。
+故dp[i][i]=0。
+
+*/
+
+public class Solution {
+    /**
+     * @param n an integer
+     * @return how much money you need to have to guarantee a win
+     */
+    public int getMoneyAmount(int n) {
+        int[][] dp = new int[n + 1][n + 1];
+        
+        for (int len = 2; len <= n; len++) {
+            for (int i = 1; i <= n - len + 1; i++) {
+                int min = Integer.MAX_VALUE;
+                min = i + dp[i + 1][i + len - 1];
+                min = Math.min(min, i + len - 1 + dp[i][i + len - 2]);
+                
+                for (int j = i + 1; j < i + len - 1; j++) {
+                    min = Math.min(min, j + Math.max(dp[i][j - 1], dp[j + 1][i + len - 1]));
+                }
+                
+                dp[i][i + len - 1] = min;
+            }
+        }
+        
+        return dp[1][n];
+    }
+}