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 = 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.

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 + 1, vector<int>(n + 1, 0));
        for (int st = n; st > 0; st--) {
            for (int ed = st + 1; ed <= n; ed++) {
                int localMin = INT_MAX;
                for (int k = st; k <= ed; k++) {
                    int cur = k + std::max(k == st ? 0 : dp[st][k-1], k == ed ? 0 : dp[k+1][ed]);
                    localMin = std::min(cur, localMin);
                }
                dp[st][ed] = localMin;
            }
        }
        return dp[1][n];
    }
};