374. Guess Number Higher or Lower
- Total Accepted: 7396
- Total Submissions: 23391
- Difficulty: Easy
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.
思路：二分查找。这里的二分查找模板可以推广。下面有2个模板，注意各自的应用场景。
代码：
这个是在左闭右闭区间[low,high]中查找元素值>=查找值的第1个元素的位置，这里也考虑high位置和查找值的大小。
1 // Forward declaration of guess API. 2 // @param num, your guess 3 // @return -1 if my number is lower, 1 if my number is higher, otherwise return 0 4 int guess(int num); 5 6 class Solution { 7 public: 8 int guessNumber(int n) { 9 int low=1,high=n,mid;10 while(low<=high){11 mid=low+(high-low)/2;12 if(guess(mid)==1){13 low=mid+1;14 }15 else{16 high=mid-1;17 }18 }19 return low;20 }21 };
这个是在左闭右开区间[low,high)中查找元素值>=查找值的第1个元素的位置，这里不考虑high位置和查找值的大小。相当于STL的函数lower_bound()。
本题有些特殊，就算high所在位置不被比较，如果排查了high之前所有位置和查找值的大小，最后low和high都停留在同1个位置，这时high虽然不被比较，但肯定就是high位置元素==查找值。例如在[1,2,3]中查找3。
1 // Forward declaration of guess API. 2 // @param num, your guess 3 // @return -1 if my number is lower, 1 if my number is higher, otherwise return 0 4 int guess(int num); 5 6 class Solution { 7 public: 8 int guessNumber(int n) { 9 int low=1,high=n,mid;10 while(low