leetcode 日记 162. Find Peak Element java python
根据题目可知,输入为:一个相邻元素不相等的数列,输出为:其中一个(上)峰值的序号。并且要求时间复杂度为logn
分析:由于题目要求时间复杂度为logn,因此不能进行全部遍历。又因为只需要找到其中的一个峰值,那么,每次对半分,便可以达到logn的复杂度。
根据对半分的思路继续想,不难发现只要确定了中间位置的数是处在上升阶段还是下降阶段,就可以确定在某一侧必有一个峰值。
往复多次,即可找出两个点,其中一个一定处于某一个峰值上。
java代码:
- public int findPeakElement(int []nums)
- {
- int left=0;
- int right=nums.length-1;
- if(right==0) return 0;
- while(left<right-1)
- {
- int mid=(left+right)/2;
- if (nums[mid]<nums[mid+1])//上升
- {
- left=mid+1;
- }
- else//下降
- {
- right=mid;
- }
- }
- return nums[left]>nums[right]?left:right;
- }
python代码:
- def findPeakElement(self, nums):
- length=len(nums)
- if length==1:
- return 0
- left=0
- right=length-1
- while left<right-1:
- mid=(left+right)/2
- #increase
- if nums[mid]<nums[mid+1]:
- left=mid+1
- #decrease
- else:
- right=mid
- return left if nums[left]>nums[right] else right
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