注意：凡是以英文出現(xiàn)的蠢正，都是題目提供的，包括答案代碼里的前幾行峦耘。
題目：
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.
希望看到這里的同學能夠先思考巾遭，最好是能夠自己寫具體的實現(xiàn)，有更好的答案可以直接在下面評論能岩。
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);
// 這里說明一下，上面的幾行代碼是題目提供的
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：這里不能用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萧福，是因為不是所有的測試都能通過捧灰，這里需要考慮數(shù)字過大導致的溢出，如果給出的數(shù)字num很大统锤，那么 end + begin超出int類型的范圍。
