From db63c022e94975c1c5180a1d7383c052c27ad159 Mon Sep 17 00:00:00 2001
From: ShusenTang <tangshusen@pku.edu.cn>
Date: Mon, 16 Dec 2019 20:17:52 +0800
Subject: [PATCH] Create 375. Guess Number Higher or Lower II.md

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+# [375. Guess Number Higher or Lower II](https://leetcode.com/problems/guess-number-higher-or-lower-ii/)
+
+# 思路
+
+猜数游戏，若猜的x且猜错了则需要付出x元钱，问需要多少钱能保证猜对。即最坏情况下，至少需要花多少钱才能猜对。
+根据提示我们发现可以根据动态规划来做，而且是个区间dp，我们设
+```
+dp[i][j]为在[i, j]内做猜数游戏的话计算出的结果, 最后返回dp[1][n]
+```
+来考虑一下初始值，设想一下我们在[4,5]之间（那就只有4和5两个数了）做猜数游戏，最坏情况下至少需要花多少钱呢，很明显是4块钱，所以
+```
+初始值: dp[i][i+1] = i, dp[i][i] = 0
+```
+那么转移方程呢？先考虑一下如果在[i, j]内做猜数游戏，那么为了最后尽可能少花钱，那么我们第一次应该猜哪个数（设为k）呢？
+我们可以穷举k属于[i+1, j-1]来看k取何值时会让最终总花费最少，即计算dp[i][j]的转移方程为
+```
+for all k in [i+1, j-1]:
+  dp[i][j] = min(dp[i][j], k + max(dp[i][k-1], dp[k+1][j]));
+```
+
+由上面转移方程看出，**我们应该反向枚举i而正向枚举j**。
+
+时间复杂度O(n^3), 空间复杂度O(n^2)。
+
+本题时间复杂度可优化至O(n^2)，可参考[1](https://artofproblemsolving.com/community/c296841h1273742)和[2](https://leetcode.com/problems/guess-number-higher-or-lower-ii/discuss/84823/Java-O(n2)-DP-solution-with-clear-explanation)。
+亲测确实快一些，但是乍一看没怎么看懂，留个坑吧。
+
+# C++
+``` C++
+class Solution {
+public:
+    int getMoneyAmount(int n) {
+        vector<vector<int>>dp(n+1, vector<int>(n+1, INT_MAX));
+        
+        for(int i = n; i >= 1; i--){
+            dp[i-1][i] = i-1; dp[i][i] = 0;
+            for(int j = i+2; j <= n; j++)
+                for(int k = i+1; k < j; k++)                    
+                    dp[i][j] = min(dp[i][j], k + max(dp[i][k-1], dp[k+1][j]));
+        }
+        
+        return dp[1][n];
+    }
+};
+```
+