Guess Number Higher or Lower II -- LeetCode

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 = , I pick .
 
 First round:  You guess , I tell you that it's higher. You pay $5.
 Second round: You guess , I tell you that it's higher. You pay $7.
 Third round:  You guess , I tell you that it's lower. You pay $9.
 
 Game over.  is the number I picked.
 
 You end up paying $ + $ + $ = $.

Given a particular n ≥ 1, find out how much money you need to have to guarantee a win.

思路：DP。

设OPT(i, j)表示i到j范围内的最优解。假设我们猜的数字是k，则i <= k <= j。之后，k将该问题分解为两个子问题，OPT(i, k-1)和OPT(k+1, j)。我们要的是其中的最坏情况。因此OPT(i, j) = min(k + max(OPT(i, k-1), OPT(k+1, j)))

复杂度为O(n^3).

 class Solution {
 public:
     int getMoneyAmount(int n) {
         vector<vector<int> > dp(n + , vector<int>(n + , ));
         for (int st = n; st > ; st--) {
             for (int ed = st + ; ed <= n; ed++) {
                 int localMin = INT_MAX;
                 for (int k = st; k <= ed; k++) {
                     int cur = k + std::max(k == st ?  : dp[st][k-], k == ed ?  : dp[k+][ed]);
                     localMin = std::min(cur, localMin);
                 }
                 dp[st][ed] = localMin;
             }
         }
         return dp[][n];
     }
 };