LeetCode*374. Guess Number Highe
2017-07-18 本文已影响0人
_Xie_
注意:凡是以英文出现的,都是题目提供的,包括答案代码里的前几行。
题目:
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.
希望看到这里的同学能够先思考,最好是能够自己写具体的实现,有更好的答案可以直接在下面评论。
答案:
// 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);
// 这里说明一下,上面的几行代码是题目提供的
class Solution {
public:
int guessNumber(int n) {
int begin = 0;
int end = n;
int num, res;
while (begin <= end) {
num = (end - begin)/2 + begin; // 注意:这里不能用num = (end + begin)/2
res = guess(num);
if (res == 0) {
return num;
}
else if (res == 1) {
begin = num + 1;
}
else {
end = num - 1;
}
}
return 0;
}
};
解析:
不能用 num = (end + begin) / 2
,是因为不是所有的测试都能通过,这里需要考虑数字过大导致的溢出,如果给出的数字num
很大,那么 end + begin
超出int
类型的范围。