LeetCode Guess Number Higher or Lower II(动态规划)

问题：从1到 n之间选取一个数字，如果猜了数字x并且猜错了，需要支付金额为x的现金。问至少需要支付多少

思路：用dp(i,j)表示i到j之间时的最小值。则有dp(i,j)=min{pivot + max(dp(i,pivot-1), dp(pivot +1, j))，pivot属于[i,j]

具体代码参考：

https://github.com/wuli2496/OJ/tree/master/LeetCode/Guess%20Number%20Higher%20or%20Lower%20II

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