【LeetCode & 劍指offer 刷題筆記】目錄（持續更新中...）

Find Peak Element

A peak element is an element that is greater than its neighbors. Given an input array   nums , where   nums[i] ≠ nums[i+1] , find a peak element and return its index. The array may contain multiple peaks, in that case return the index to any one of the peaks is fine. You may imagine that   nums[-1] = nums[n] = -∞ . Example 1: Input: nums = [1,2,3,1] Output: 2 Explanation: 3 is a peak element and your function should return the index number 2. Example 2: Input: nums = [ 1,2,1,3,5,6,4] Output: 1 or 5 Explanation: Your function can return either index number 1 where the peak element is 2, or index number 5 where the peak element is 6. Note: Your solution should be in logarithmic complexity.

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C++   //問題：找極大值元素（返回任意一個極大值均可） /* 方法一：找最大值一定是極大值 O(n) */ class Solution { public :     int findPeakElement ( vector < int >& nums )     {         int n = nums . size ();         if ( n == 0 ) return - 1 ; //為空時，返回-1         int max_index = 0 ;         for ( int i = 0 ; i < n ; i ++) //找序列中的最大值，一定是極大值         {             if ( nums [ i ]> nums [ max_index ]) max_index = i ;         }         return max_index ;     } };   /* 掌握 方法二：藉助二分查詢思路 right指的數比後一個數大，left指的數比前一個數大，當兩個“指標”相遇時，就可以滿足極大值條件（兩個指標按二分跨度走，故效率較高） 有點像lower_bound函式 O(logn) */ class Solution { public :     int findPeakElement ( vector < int >& nums )     {         if ( nums . empty () return - 1 ; // 為空時，返回 -1                  int left = 0 , right = nums . size ()- 1 ;         while ( left < right ) //left = right 時退出迴圈，結果是 left,right,mid 均指向極大值位置         {             int mid = ( left + right )/ 2 ;             if ( nums [ mid ] < nums [ mid + 1 ]) //如果中間值比右邊值小，說明峰值在右邊                 left = mid + 1 ;             else // 如果中間值比右邊值大，說明峰值在左邊（包括 a[mid] ，故取 right=mid ）                 right = mid ;         }                return right;            } };