162. Find Peak Element的C++解法
看到時間複雜度要求為對數立刻想到二分法。注意有一個邊界條件,在LintCode中的尋找峰值也有一個類似條件,本質是一樣的,即中間的數兩邊,增長的一端一定有峰值。
class Solution { public: int findPeakElement(vector<int>& nums) { if (nums.size()==1) return 0; int head=0; int tail=nums.size()-1; while (head<tail) { int mid=(head+tail)/2; if (nums[mid+1]>nums[mid]) head=mid+1; else if (nums[mid-1]>nums[mid]) tail=mid-1; else return mid; } return head; } };
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