LeetCode:数组峰值与谷值问题总结
问题描述:
(1)剑指 Offer 11. 旋转数组的最小数字 , 存在重复元素找最小值
(2)LeetCode: 153. 寻找旋转排序数组中的最小值,元素不相同,找最下值
(3)LeetCode: 154. 寻找旋转排序数组中的最小值 II,存在重复元素,找最小值
(4)LeetCode: 162. 寻找峰值,无序,求峰值
(5)LeetCode: 852. 山脉数组的峰顶索引, 找索引
(1)剑指 Offer 11. 旋转数组的最小数字 Python3实现:
python
class Solution:
def minArray(self, numbers: List[int]) -> int:
low, high = 0, len(numbers) - 1
while low < high:
pivot = low + (high - low) // 2
if numbers[pivot] < numbers[high]:
high = pivot
elif numbers[pivot] > numbers[high]:
low = pivot + 1
else:
high -= 1
return numbers[low]
(2)LeetCode: 153. 寻找旋转排序数组中的最小值 Python3实现:
python
class Solution:
def findMin(self, nums: List[int]) -> int:
left, right = 0, len(nums) - 1 # 左闭右闭区间,如果用右开区间则不方便判断右值
while left < right: # 循环不变式,如果left == right,则循环结束
mid = (left + right) >> 1 # 地板除,mid更靠近left
if nums[mid] > nums[right]: # 中值 > 右值,最小值在右半边,收缩左边界
left = mid + 1 # 因为中值 > 右值,中值肯定不是最小值,左边界可以跨过mid
elif nums[mid] < nums[right]: # 明确中值 < 右值,最小值在左半边,收缩右边界
right = mid # 因为中值 < 右值,中值也可能是最小值,右边界只能取到mid处
return nums[left] # 循环结束,left == right,最小值输出nums[left]或nums[right]均可
(3)LeetCode: 154. 寻找旋转排序数组中的最小值 II Python3实现:
python
class Solution:
def findMin(self, nums: List[int]) -> int:
low, high = 0, len(nums) - 1
while low < high:
pivot = low + (high - low) // 2
if nums[pivot] < nums[high]:
high = pivot
elif nums[pivot] > nums[high]:
low = pivot + 1
else:
high -= 1
return nums[low]
(4)LeetCode: 162. 寻找峰值 Python3实现:
python
class Solution:
def findPeakElement(self, nums):
low, high = 0, len(nums) - 1
while low < high:
mid = (low + high) // 2
if nums[mid] > nums[mid + 1]:
high = mid
else:
low = mid + 1
return low
(5)LeetCode: 852. 山脉数组的峰顶索引 Python3实现:
python
class Solution:
def peakIndexInMountainArray(self, arr: List[int]) -> int:
n = len(arr)
low = 0
high = n - 1
while low < high:
mid = (low + high) // 2
if arr[mid] > arr[mid+1]:
high = mid
else:
low = mid + 1
return low
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声明: 总结学习,有问题或不当之处,可以批评指正哦,谢谢。