코딩테스트/Programmers
[Leet Code] 374. Guess Number Higher or Lower
딥런
2022. 1. 28. 03:12
Same Tree
문제 설명
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.
제한 사항
- 1 <= n <= 231 - 1
- 1 <= pick <= n
입출력 예
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
Python 코드
Python code
# The guess API is already defined for you.
# @param num, your guess
# @return -1 if my number is lower, 1 if my number is higher, otherwise return 0
# def guess(num: int) -> int:
class Solution:
def guessNumber(self, n: int) -> int:
left, right = 1, n
while True:
middle_num = int((left + right) / 2)
if guess(middle_num) == 1:
left = middle_num + 1
elif guess(middle_num) == -1:
right = middle_num - 1
else:
return middle_num
* 참고 링크 : https://somjang.tistory.com/entry/leetCode-374-Guess-Number-Higher-or-Lower-Python
C++ 코드
C ++ code
//Runtime: 0 ms, faster than 100.00% of C++ online submissions for Guess Number Higher or Lower.
//Memory Usage: 7.3 MB, less than 100.00% of C++ online submissions for Guess Number Higher or Lower.
// 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 left = 1, right = n, cur = left+(right-left)/2;
while(true){
switch(guess(cur)){
case -1:
right = cur-1;
cur = left+(right-left)/2;
break;
case 1:
left = cur+1;
cur = left+(right-left)/2;
break;
case 0:
return cur;
}
}
return 0;
}
};
출처