Find Peak Element.png
===================== 解題思路 =====================
二分法直到找出 A[mid] > A[mid - 1] && A[mid] > A[mid + 1] 為止
如果找到的點比前一個點小 (A[mid] < A[mid -1]) 說明可能的 peak 在左邊 所以設定 right = mid 來往左邊找 反之則是設定 left = mid 往右邊找
===================== C++ code ====================
<pre><code>
class Solution {
public:
/**
* @param A: An integers array.
* @return: return any of peek positions.
*/
int findPeak(vector<int> A) {
// write your code here
int left = 0, right = A.size() - 1;
while(left + 1 < right)
{
int mid = left + (right - left) / 2;
if(mid > 0 && A[mid] > A[mid - 1] && A[mid] > A[mid + 1]) return mid;
if(A[mid] < A[mid -1]) right = mid;
else left = mid;
}
if(A[left] > A[right]) return left;
else return right;
}
};
<code><pre>
