|
| 1 | +""" |
| 2 | +A peak element is an element that is greater than its neighbors. |
| 3 | + |
| 4 | +Given an input array nums, where nums[i] ≠ nums[i+1], find a peak element and return its index. |
| 5 | + |
| 6 | +The array may contain multiple peaks, in that case return the index to any one of the peaks is fine. |
| 7 | + |
| 8 | +You may imagine that nums[-1] = nums[n] = -∞. |
| 9 | + |
| 10 | +Example 1: |
| 11 | + |
| 12 | +Input: nums = [1,2,3,1] |
| 13 | +Output: 2 |
| 14 | +Explanation: 3 is a peak element and your function should return the index number 2. |
| 15 | +Example 2: |
| 16 | + |
| 17 | +Input: nums = [1,2,1,3,5,6,4] |
| 18 | +Output: 1 or 5 |
| 19 | +Explanation: Your function can return either index number 1 where the peak element is 2, |
| 20 | + or index number 5 where the peak element is 6. |
| 21 | +Note: |
| 22 | + |
| 23 | +Your solution should be in logarithmic complexity. |
| 24 | + |
| 25 | +log n待续。 |
| 26 | + |
| 27 | +先O(n)。 |
| 28 | + |
| 29 | +看来测试数据不多,O(n) 的可以 beat 100%. |
| 30 | + |
| 31 | +测试地址: |
| 32 | +https://leetcode.com/problems/find-peak-element/description/ |
| 33 | + |
| 34 | +""" |
| 35 | +class Solution(object): |
| 36 | + def findPeakElement(self, nums): |
| 37 | + """ |
| 38 | + :type nums: List[int] |
| 39 | + :rtype: int |
| 40 | + """ |
| 41 | + |
| 42 | + length = len(nums) |
| 43 | + |
| 44 | + for i in range(1, length-1): |
| 45 | + if nums[i-1] < nums[i] > nums[i+1]: |
| 46 | + return i |
| 47 | + |
| 48 | + if length <= 2: |
| 49 | + return nums.index(max(nums)) |
| 50 | + else: |
| 51 | + if nums[0] > nums[1]: |
| 52 | + return 0 |
| 53 | + elif nums[-1] > nums[-2]: |
| 54 | + return length-1 |
| 55 | + return None |
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