Leetcode 374 & 375

Leetcode 374 & 375

374 Guess Number Higher or Lower

Description

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!

题意分析

374是一道简单题，给出一个数字n，猜1到n中的数字中的一个，通过调用API guess函数来判断你的猜测是比结果大还是小，返回1表示猜得太大，-1反之，0就正确了，通过一个简单的二分查找就可以得出答案，solution里面还有用三分查找，道理一样的。二分的代码如下:

// 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 low = 1, high = n, mid, res;
        while (low <= high) {
            mid = low + (high - low) / 2;
            res = guess(mid);
            if (res == 0)
                return mid;
            if (res < 0)
                high = mid - 1;
            else
                low = mid + 1;
        }
        return -1;
    }
};

375 Guess Number Higher or Lower II

Description

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.

题意分析

这道题就比较难了，题目意思是这样的，同样是猜1到n的一个数字，比如说n=10，结果为8，我猜5，错了，我要给你5块钱，你说猜小了。我再猜9，错了，我给你9块钱，你说猜大了。我再猜8，猜对了。一路下来我要给17块钱，问题就是求解我至少需要多少钱，才能保证我一定能赢？
解题之前先再来思考一下题目的情况，大致就是“要找出最小的最大值”，就是从1到n中，我应该怎么选才能使得我花的钱最少，但是又能保证我赢。比如说n=5，结果为2，如果我一开始猜3，大了，然后我可以选择猜1、2或者4、5，那么这个时候我就应该至少有3 + 4 = 7块钱才能保证我一定能赢，因为如果我只有3 + 2 = 5块钱，并且结果为4的情况下，我第二轮选4或者5都有可能让我输，但是如果我有4，无论我选1、2、4，我都能保证最后我能赢。然后再讨论一开始选择1、2、3、4、5五种情况下的花费最少的情况，就是结果。
想到这里，就可以用分治(动态规划)的方法来做，先选出一个i，然后把一个数组分成i的左边跟右边，找出左边右边的较大者加上i，然后取出1到n中最小的结果的那个i的方案，就是最后的答案,代码如下：

class Solution {
public:
        // 数组nums[i][j]表示的是在已经选择了i的前提下，选中最后结果为j的情况所需的总的花费
    int getMoneyAmount(int n) {
        int **nums = new int*[n + 1];
        for (int i = 0; i <= n; i++) {
            nums[i] = new int [n + 1];
        }
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
                nums[i][j] = 0;
        int res = solve(nums, 1, n);
        for (int i = 0; i <= n; i++)
            delete []nums[i];
        delete []nums;
        return res;
    }
    int solve(int** nums, int left, int right) {
        // 当左边的边界大于等于右边边界的时候，表明已经到了极限，游戏结束
        if (left >= right)
            return 0;
        // 这里用一个保存的方法，避免重新进入数组的时候重新计算
        if (nums[left][right] != 0)
            return nums[left][right];
        int res = INT_MAX;
        int temp;
        for (int i = left; i <= right; i++) {
            // 考虑最坏的情况
            temp = i + max(solve(nums, left, i - 1), solve(nums, i + 1, right));
            if (temp < res)
                res = temp;
        }
        // 记得要把数组的值保存
        nums[left][right] = res;
        return res;
    }
};