LeetCode //C – 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 will tell you whether the number I picked is higher or lower than your guess.

You call a pre-defined API int guess(int num), which returns three possible results:

• -1: Your guess is higher than the number I picked (i.e. num > pick).
• 1: Your guess is lower than the number I picked (i.e. num < pick).
• 0: your guess is equal to the number I picked (i.e. num == pick).

Return the number that I picked.
 

Example 1:

Input: n = 10, pick = 6
Output: 6

Example 2:

Input: n = 1, pick = 1
Output: 1

Example 3:

Input: n = 2, pick = 1
Output: 1

Constraints:

• $1<=n<=2^{31}-1$
• $1<=pick<=n$

From: LeetCode
Link: 374. Guess Number Higher or Lower

Solution:

Ideas：

• We initialize two pointers, left and right, to the beginning and end of the range, respectively.
• We perform a binary search by repeatedly choosing the middle element of the current range as our guess.
• We call the guess API with our current guess. If the return value is 0, we have found the correct number and return it. If the return value is -1, our guess is too high, so we adjust the right boundary. If the return value is 1, our guess is too low, so we adjust the left boundary.
• This process continues until we find the correct number.

Code:

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

int guessNumber(int n) {
    long long left = 1, right = n;
    while (left <= right) {
        // Use long long for mid to avoid integer overflow
        long long mid = left + (right - left) / 2;
        int res = guess(mid);
        if (res == 0) {
            // The guess is correct
            return mid;
        } else if (res < 0) {
            // The picked number is lower than our guess
            right = mid - 1;
        } else {
            // The picked number is higher than our guess
            left = mid + 1;
        }
    }
    // This line should never be reached if the API behaves as expected
    return -1;
}

代码007普通

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