今天，试着做了一下LeetCode OJ上面的第375道题：Guess Number Higher or Lower II
原题链接：https://leetcode.com/problems/guess-number-higher-or-lower-ii/
具体题目如下：
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.
上网搜了一下，了解到用动态规划的思路解决。
这时，有两种思路（猜的数范围为：1-n）：
dp[p][q]表示猜数的范围在p-q之间要保证赢需要的最小花费。此题为：dp[1][n].
思路一：
假设我们随意猜的数为k,
dp[1][n] = k + max{dp[1][k-1],dp[k+1][n]},
如此，最后得到了一个值，此值为在1-n区间猜数保证赢所需要的最小花费（最大花费？）。
在此过程中，我们不断的在寻求每个区间的最大花费，这样会不会导致所猜的数一直在改变，感觉这样求出的最大就太大了，而且也没有什么意义。
默默的感觉是自己想错了，所猜的数没有在变？
还有一个问题是：假设所猜的数不变，那么如果一直取子区间所需要的最大花费，那么最小又从哪里体现呢？
思路二：
在1-n个数里面，我们任意猜一个数(设为i)，保证获胜所花的钱应该为 i + max(w(1 ,i-1), w(i+1 ,n))，这里w(x,y))表示猜范围在(x,y)的数保证能赢应花的钱，则我们依次遍历 1-n作为猜的数，求出其中的最小值即为答案，即最小的最大值问题。
