Follow up for "Find Minimum in Rotated Sorted Array":
What if duplicates are allowed?Would this affect the run-time complexity? How and why?
Suppose a sorted array is rotated at some pivot unknown to you beforehand.
(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).
Find the minimum element.
The array may contain duplicates.
153. Find Minimum in Rotated Sorted Array 的拓展，这个题里允许有重复的元素。原来是依靠中间和边缘元素的大小关系，来判断哪一半是有序的。而现在因为重复元素的出现，如果遇到中间和边缘相等的情况，就无法判断哪边有序，因为哪边都有可能有序。假设原数组是{1,2,3,3,3,3,3}，那么旋转之后有可能是{3,3,3,3,3,1,2}，或者{3,1,2,3,3,3,3}，判断左边缘和中心的时候都是3，就不知道应该截掉哪一半。解决的办法是对边缘移动一步，直到边缘和中间不在相等或者相遇，这就导致了会有不能切去一半的可能。所以最坏情况就会出现每次移动一步，总共移动n，算法的时间复杂度j: O(logn) ~ O(n)。
Java:
public int findMin(int[] num) {
if(num == null || num.length==0)
return 0;
int l = 0;
int r = num.length-1;
int min = num[0];
while(l<r-1)
{
int m = (l+r)/2;
if(num[l]<num[m])
{
min = Math.min(num[l],min);
l = m+1;
}
else if(num[l]>num[m])
{
min = Math.min(num[m],min);
r = m-1;
}
else
{
l++;
}
}
min = Math.min(num[r],min);
min = Math.min(num[l],min);
return min;
}
Python:
class Solution(object):
def findMin(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
left, right = 0, len(nums) - 1
while left < right:
mid = left + (right - left) / 2
if nums[mid] == nums[right]:
right -= 1
elif nums[mid] < nums[right]:
right = mid
else:
left = mid + 1
return nums[left]
Python：
class Solution2(object):
def findMin(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
left, right = 0, len(nums) - 1
while left < right and nums[left] >= nums[right]:
mid = left + (right - left) / 2
if nums[mid] == nums[left]:
left += 1
elif nums[mid] < nums[left]:
right = mid
else:
left = mid + 1
return nums[left]
C++:
class Solution {
public:
int findMin(vector<int> &nums) {
if (nums.empty()) return 0;
int left = 0, right = nums.size() - 1, res = nums[0];
while (left < right - 1) {
int mid = left + (right - left) / 2;
if (nums[left] < nums[mid]) {
res = min(res, nums[left]);
left = mid + 1;
} else if (nums[left] > nums[mid]) {
res = min(res, nums[right]);
right = mid;
} else ++left;
}
res = min(res, nums[left]);
res = min(res, nums[right]);
return res;
}
};
类似题目：
[LeetCode] 33. Search in Rotated Sorted Array 在旋转有序数组中搜索
[LeetCode] 81. Search in Rotated Sorted Array II 在旋转有序数组中搜索 II
[LeetCode] 153. Find Minimum in Rotated Sorted Array 寻找旋转有序数组的最小值
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