做leetcode的题
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;
第一种计算方法会Time Limit Exceeded，原因可能是（low + high）的结果超过int的最大范围，越界。
可以使用第一种公式，但把数据改成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；
然后注意审题，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:
- 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.
- Take a small example (n = 3). What do you end up paying in the worst case?
- Check out this article if you're still stuck.
- The purely recursive implementation of minimax would be worthless for even a small n. You MUST use dynamic programming.
- 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：最大值最小化算法（更新：用最小化的方法（这里是二分法），找到所有情况，取最大值，就是题目中需要的“至少”）；也就是求的是最大值，但是是最大值中的最小的那一个；那么逻辑应该是很清晰的，两步，找到最大值，再找最大值的最小值；
基本实现：递归；——这里特别吊的是别个用了个二维数组来比较，用二维数组的序号/位置，来分析n个数的情况——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;
}
}
这个递归用的我心服口服。。。。