# [LeetCode] 375. Guess Number Higher or Lower II 猜数字大小之二

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.

Hint:

1. The best strategy to play the game is to minimize the maximum loss you could possibly face. Another strategy is to minimize the expected loss. Here, we are interested in thefirst scenario.
2. Take a small example (n = 3). What do you end up paying in the worst case?
4. The purely recursive implementation of minimax would be worthless for even a small n. You MUST use dynamic programming.
5. As a follow-up, how would you modify your code to solve the problem of minimizing the expected loss, instead of the worst-case loss?

Credits:
Special thanks to @agave and @StefanPochmann for adding this problem and creating all test cases.

class Solution {
public:
int getMoneyAmount(int n) {
vector<vector<int>> dp(n + 1, vector<int>(n + 1, 0));
for (int i = 2; i <= n; ++i) {
for (int j = i - 1; j > 0; --j) {
int global_min = INT_MAX;
for (int k = j + 1; k < i; ++k) {
int local_max = k + max(dp[j][k - 1], dp[k + 1][i]);
global_min = min(global_min, local_max);
}
dp[j][i] = j + 1 == i ? j : global_min;
}
}
return dp[1][n];
}
};

class Solution {
public:
int getMoneyAmount(int n) {
vector<vector<int>> memo(n + 1, vector<int>(n + 1, 0));
return helper(1, n, memo);
}
int helper(int start, int end, vector<vector<int>>& memo) {
if (start >= end) return 0;
if (memo[start][end] > 0) return memo[start][end];
int res = INT_MAX;
for (int k = start; k <= end; ++k) {
int t = k + max(helper(start, k - 1, memo), helper(k + 1, end, memo));
res = min(res, t);
}
return memo[start][end] = res;
}
};

Github 同步地址：

https://github.com/grandyang/leetcode/issues/375

Guess Number Higher or Lower

Find K Closest Elements

https://leetcode.com/problems/guess-number-higher-or-lower-ii/

https://leetcode.com/problems/guess-number-higher-or-lower-ii/discuss/84787/Java-DP-solution

https://leetcode.com/problems/guess-number-higher-or-lower-ii/discuss/84764/Simple-DP-solution-with-explanation~~

LeetCode All in One 题目讲解汇总(持续更新中...)

posted @ 2016-07-16 23:48  Grandyang  阅读(22189)  评论(10编辑  收藏  举报