leetcode 374. Guess Number Higher or Lower

374. Guess Number Higher or Lower

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.

标准二分查找

// Forward declaration of guess API.
// @param num, your guess
// @return -1 if my number is lower, 1 if my number is higher, otherwise return 0
int guess(int num);

class Solution {
public:
    int guessNumber(int n) 
    {
        int start = 1, end = n;
        while (start + 1 < end)
        {
            int mid = (end - start) / 2 + start;
            int k = guess(mid);
            if (k == 0)
                return mid;
            else if (k == -1)
                end = mid;
            else
                start = mid;
        }
        return (guess(start) == 0) ? start : end;  //double check
    }
};

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.

1、使用记忆化搜索，保存已经遍历过的 段。

2、开始将ret = min(ret, i + max(helper(start, i - 1), helper(i + 1, end)));写成了ret = min(ret, i + helper(start, i - 1) + helper(i + 1, end));非常不应该。

这个题在选了 i 之后，就只会往一边发展，所以就找花销大的那边就够了。

class Solution {
public:
    int getMoneyAmount(int n)
    {
        count = vector<vector<int>> (n + 1, vector<int> (n + 1, -1));
        return helper(1, n);
    }
private:
    vector<vector<int>> count;
    
    int helper(int start, int end)    //helper代表在[start, end]中保证能找到一个值的 最小代价
    {
        if (start > end) return 0;
        if (count[start][end] != -1)  //说明已经记忆化搜索过这一段
            return count[start][end];
        else if (start == end)
        {
            count[start][end] = 0;
            return 0;
        }
        else
        {
            int ret = INT_MAX;
            for (int i = start; i <= end; i++)
            {
                // i(当前代价） + 往后面选的最大代价（可能往左或者是右）
                ret = min(ret, i + max(helper(start, i - 1), helper(i + 1, end)));
            }
            count[start][end] = ret;
            return ret;
        }
    }
};

int main( )
{
    
    return 0;
}