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:
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。。。。
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