题目要求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: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.一个猜数字游戏，数字区间为1~n，每猜一次，会有人告诉你猜中了或者当前的数字是大于结果值还是小于结果值。猜对则本次猜测免费，猜错则本次猜测需要花费和数字等额的金钱。问如果要确保能够猜中数字，最少要花费多少钱。其实这题的英文表述有些问题，确切来说，在所有能够确保找到目标值的方法中，找到花费金钱最少的哪种。思路我们先用一个比较简单的数字来找规律。当n等于3时，即从1,2,3中找到目标数字，确保找到一个数字至少需要多少钱。查询序列如下：目标值3：1,2 2目标值2：1,3目标值1：3,2,2可见如果要找到1，2，3中任意一个数字，最少要花费2元钱，即从2开始查询，如果命中，则花费0元，如果没有命中也知道目标值比2小还是比2大，下次猜测一定命中，因此该序列中找到任何一个数字最多花费2元钱。如此可见，假如要知道min(1,n)的值，只要找到花费最小的中间点k，即递归的公式相当于min(1,n) = k + Math.max(min(1, k-1), min(k+1,n)) 1<=k<=n找到最小的min即可。思路一：自顶向下的动态规划 public int getMoneyAmount(int n) { int[][] tmp = new int[n+1][n+1]; for(int[] row: tmp){ Arrays.fill(row, Integer.MAX_VALUE); } return getMoneyAmount(n, 1, n, tmp); } public int getMoneyAmount(int n, int lft, int rgt, int[][] tmp) { if(lft>=rgt) return 0; if(tmp[lft][rgt] != Integer.MAX_VALUE) return tmp[lft][rgt]; for(int i = lft ; i<=rgt ; i++) { tmp[lft][rgt] = Math.min(tmp[lft][rgt], Math.max(i + getMoneyAmount(n, lft, i-1, tmp), i + getMoneyAmount(n, i+1, rgt, tmp))); } return tmp[lft][rgt]; }思路二：自底向上的动态规划 public int getMoneyAmount(int n) { int[][] dp = new int[n+1][n+1]; for(int i = 2 ; i<=n ; i++) { for(int j = i-1 ; j>0 ; j–) { int min = Integer.MAX_VALUE; for(int k = j+1 ; k<i ; k++) { min = Math.min(min, k + Math.max(dp[j][k-1], dp[k+1][i])); } dp[j][i] = j + 1 == i ? j : min; } } return dp[1][n]; }