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# 374&375. Guess Number Higher or Lower 1&2

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 is higher or lower.

You call a pre-defined API `guess(int num)` which returns 3 possible results (`-1``1`, or `0`):

`-1 : My number is lower 1 : My number is higher 0 : Congrats! You got it!`

Example:

`n = 10, I pick 6.Return 6.基本就是这样：关键在于怎么找，怎么去guess；基本点：查找/随机；（参数）递归；进阶点：TLE错误解决——`

mid = (low + high) / 2;
mid = low + (high - low) / 2;

`/* The guess API is defined in the parent class GuessGame.   @param num, your guess   @return -1 if my number is lower, 1 if my number is higher, otherwise return 0      int guess(int num); */public class Solution extends GuessGame {    public int guessNumber(int n) {        if (guess(n)== 0) return n;        int left=0;        int right=n;        while (left<right) {          int mid=left+(right-left)/2;          int re=guess(mid);          if (re==0){              return mid;          }else if(guess(mid)==-1){            right=mid;          }else{            left=mid;          }         // return left;        }        return left;    }}`

375. Guess Number Higher or Lower II

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?

`public class Solution {  public int getMoneyAmount(int n) {    int[][] table = new int[n + 1][n + 1];  //0    return payForRange(table, 1, n);  }  //return the amount paid for the game within range [start,end]  private int payForRange(int[][] dp, int start, int end) {    if (start >= end)      return 0;    if (dp[start][end] != 0)      return dp[start][end];    int minimumForCurrentRange = Integer.MAX_VALUE;    for (int x = start; x <= end; ++x) {      //calculate the amount to pay if pick x.      int pay = x + Math.max(payForRange(dp, start, x - 1), payForRange(dp, x + 1, end));      //calculate min of maxes      minimumForCurrentRange = Math.min(minimumForCurrentRange, pay);    }    dp[start][end] = minimumForCurrentRange;    return minimumForCurrentRange;  }}`

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