Description
Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:
- [4,5,6,7,0,1,2] if it was rotated 4 times.
- [0,1,2,4,5,6,7] if it was rotated 7 times.
Notice that rotating an array [a[0], a[1], a[2], …, a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], …, a[n-2]].
Given the sorted rotated array nums of unique elements, return the minimum element of this array.
You must write an algorithm that runs in O(log n) time.
Examples
Example 1:
Input: nums = [3,4,5,1,2]
Output: 1
Explanation: The original array was [1,2,3,4,5] rotated 3 times.
Example 2:
Input: nums = [4,5,6,7,0,1,2]
Output: 0
Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.
Example 3:
Input: nums = [11,13,15,17]
Output: 11
Explanation: The original array was [11,13,15,17] and it was rotated 4 times.
Constraints:
n == nums.length
1 <= n <= 5000
-5000 <= nums[i] <= 5000
All the integers of nums are unique.
nums is sorted and rotated between 1 and n times.
思路
这题是想要找出循环升序链表里面最小的那个数字，题目也说了要用O(log n)的时间复杂度，所以肯定是用二分法
所以问题的关键就在于怎么设定二分规则
这里的二分规则就是要明确再给出一个循环升序链表的时候，那个最小值会位于什么地方
这里以最小值在middle左侧为例
- 首先第一种也是最好找的，就是nums[start] < nums[end]
因为是循环升序，所以当循环起来的时候，必然有nums[start] > nums[end]，而当不满足这个条件的时候，就是他没有循环，只是升序排列的情况，那最小值必然在左侧，甚至就是nums[start] - 第二种是nums[middle] < nums[start]
在普通升序情况下，nums[start] < nums[middle]是正常情况，当不满足这个条件的时候，意味着循环发生在左侧的环境内，也就是最小值在左侧产生
那么这两种情况以外的其他情况，就是最小值在右侧的情况啦
代码
class Solution {
public int min(int[] nums, int start, int end){
if(start == end)
return nums[start];
int mid = (start + end) / 2;
if(nums[start] < nums[end] || nums[start] > nums[mid])
return min(nums, start, mid);
else
return min(nums, mid + 1, end);
}
public int findMin(int[] nums) {
int len = nums.length - 1;
return min(nums, 0, len);
}
}
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