leetcode 375. Guess Number Higher or Lower II 解题报告
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原题链接
原题链接
解题思路
刚开始并不是很理解题意,然后去看了看讨论区的大神们。大概明白了。
对于每一个数字x在数组[i~j],我们算他的花费为 x + max{DP([i~x-1]), DP([x+1, j])}。
the max意思是无论你猜哪个数字,这个数字的反馈都是最差的,需要花费很多。
the min是指在这么多最差反馈里挑出最少的花费。
数组dp[i][j] 表示i~j的minmax花费。
解题代码
public class Solution { public int getMoneyAmount(int n) { int[][] dp = new int[n+1][n+1]; return DP(dp,1,n); } public int DP(int[][] dp,int low,int high) { if(high <= low) { return 0; } if(dp[low][high] != 0) { return dp[low][high]; } int res = Integer.MAX_VALUE; for (int i = low;i <= high;i++) { int tmp = i + Math.max(DP(dp,low,i-1),DP(dp,i+1,high)); res = Math.min(tmp,res); } dp[low][high] = res; return res; }}
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