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## Problem

Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,4,4,5,6,7] might become:

• [4,5,6,7,0,1,4] if it was rotated 4 times.
• [0,1,4,4,5,6,7] if it was rotated 7 times.

Notice that rotating an array [a, a, a, …, a[n-1]] 1 time results in the array [a[n-1], a, a, a, …, a[n-2]].

Given the sorted rotated array nums that may contain duplicates, return the minimum element of this array.

You must decrease the overall operation steps as much as possible.

Example 1:

Input: nums = [1,3,5]
Output: 1


Example 2:

Input: nums = [2,2,2,0,1]
Output: 0


## Code

class Solution {
public int findMin(int[] nums) {
if(nums == null || nums.length == 0){
return -1;
}

int start = 0;
int end = nums.length - 1;

while(start + 1 < end){
int mid = start + (end - start) / 2;
// 缩小范围
if(nums[mid] == nums[start]){
start++;
} else if (nums[mid] == nums[end]){
end--;
// 使用nums[end]作为标准
} else if (nums[mid] < nums[end]){
end = mid;
} else if (nums[mid] > nums[end]){
start = mid;
}
}

if(nums[start] < nums[end]){
return nums[start];
} else {
return nums[end];
}
}
}