75.寻找峰值-创新互联

• 相邻位置的数字是不同的
• A[0] < A[1] 并且 A[n - 2] > A[n - 1]

假定P是峰值的位置则满足A[P] > A[P-1]且A[P] > A[P+1]，返回数组中任意一个峰值的位置。

 注意事项

• It's guaranteed the array has at least one peak.
• The array may contain multiple peeks, find any of them.
• The array has at least 3 numbers in it.

样例

给出数组[1, 2, 1, 3, 4, 5, 7, 6]返回1, 即数值 2 所在位置, 或者6, 即数值 7 所在位置.

使用递归思想，每次搜索范围减半

 1   int findPeak(vector& A) {
 2 // write your code here 3 int low =0;
 4 int high=A.size();
 5 int mid;
 6 while(high>low){
 7             mid=(low+high)/2;
 8  if(A[mid]>A[mid-1] && A[mid]>A[mid+1]){
 9   return mid;
10             }
11  else if(A[mid-1]>A[mid]){
12                 high=mid;
13             }
14  else{
15                 low=mid;
16             }
17         }
18     }