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author | jack <[email protected]> | 2024-06-14 13:13:15 -0600 |
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committer | jack <[email protected]> | 2024-06-14 13:13:15 -0600 |
commit | d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa (patch) | |
tree | 96df922dae2a5073d9a50ea70c7aae6aa37e3ebe /find_peak_element.py | |
download | leetcode-d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa.tar.gz leetcode-d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa.zip |
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Diffstat (limited to 'find_peak_element.py')
-rw-r--r-- | find_peak_element.py | 34 |
1 files changed, 34 insertions, 0 deletions
diff --git a/find_peak_element.py b/find_peak_element.py new file mode 100644 index 0000000..18ef2f9 --- /dev/null +++ b/find_peak_element.py @@ -0,0 +1,34 @@ +# 162. Find Peak Element + +""" +a peak element is an element that is strictly greater than its neighbours. +given a 0 indexed array 'nums', find a peak element and return its index. if +the array contains multiple peaks, return the index to any of the peaks. you +may imagine that 'nums[-1] = nums[n] = -inf' in other words, an element is +always considered to be strictly greater than a neighbour that is outside the +array. you must write an algorithm that runs in 'O(log n)' time. +""" + + +class Solution(object): + def findPeakElement(self, nums): + """ + :type nums: List[int] + :rtype: int + """ + left, right = 0, len(nums) - 1 + while left < right - 1: + mid = (left + right) / 2 + if nums[mid] > nums[mid + 1] and nums[mid] > nums[mid - 1]: + return mid + if nums[mid] < nums[mid + 1]: + left = mid + 1 + else: + right = mid - 1 + return left if nums[left] >= nums[right] else right + + +if __name__ == "__main__": + obj = Solution() + print(obj.findPeakElement(nums=[1, 2, 3, 1])) + print(obj.findPeakElement(nums=[1, 2, 1, 3, 5, 6, 4])) |