374&375. Guess Number Higher or Lower 1&2，桑切斯

374&375. Guess Number Higher or Lower 1&2，桑切斯

做leetcode的題

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.

基本就是這樣：關鍵在於怎麼找，怎麼去guess；

基本點：查找/隨機；（參數）遞歸；
進階點：TLE錯誤解決——

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

第一種計算方法會Time Limit Exceeded，原因可能是（low + high）的結果超過int的最大范圍，越界。

可以使用第一種公式，但把數據改成long型；也可以改成mid = low/2 + high/2公式。

引用自http://blog.csdn.net/y12345678904/article/details/51898958；

然後注意審題，看清楚guess和guessNum

/* 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?
3. Check out this article if you're still stuck.
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?

基本思路：最小化最大值算法；也就是求的是最大值，但是是最大值中的最小的那一個；那麼邏輯應該是很清晰的，兩步，找到最大值，再找最大值的最小值；

基本實現：遞歸；——這裡特別吊的是別個用了個二維數組來比較，用二維數組的序號/位置，來分析n個數的情況——http://www.cnblogs.com/neweracoding/p/5679936.html；

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;
  }
}

這個遞歸用的我心服口服。。。。