题目
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.
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分析
在含有重复元素的旋转有序序列中查找最小元素。
与上一题类似，LeetCode（153） Find Minimum in Rotated Sorted Array 153题中的旋转有序数组不包含重复元素，而此题允许重复元素，增加了一点难度。
我想题目重点考察的还是沿用二分查找的方法解决，思路参考。
AC代码
class Solution {
public:
//方法一：使用stl库函数
int findMin1(vector<int>& nums) {
if (nums.empty())
return 0;
vector<int>::iterator iter = min_element(nums.begin(), nums.end());
return *iter;
}
//方法二：整个数列除一处为最大值到最小值的跳转外，为两部分的递增
int findMin2(vector<int>& nums)
{
if (nums.empty())
return 0;
if (nums.size() == 1)
return nums[0];
for (size_t i = 1; i < nums.size(); ++i)
{
if (nums[i - 1] > nums[i])
return nums[i];
}//for
//没有找到跳转元素，则序列无旋转
return nums[0];
}
int findMin(vector<int> &nums)
{
if (nums.empty())
return 0;
else if (nums.size() == 1)
return nums[0];
else{
int lhs = 0, rhs = nums.size() - 1;
while (lhs < rhs && nums[lhs] >= nums[rhs])
{
int mid = (lhs + rhs) / 2;
if (nums[lhs] > nums[mid])
rhs = mid;
else if (nums[lhs] == nums[mid])
++lhs;
else
lhs = mid + 1;
}//while
return nums[lhs];
}
}
};