Find Minimum in Rotated Sorted Array II
最后更新于：2022-04-02 01:09:12
# Find Minimum in Rotated Sorted Array II
### Source
- leetcode: [Find Minimum in Rotated Sorted Array II | LeetCode OJ](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/)
- lintcode: [(160) Find Minimum in Rotated Sorted Array II](http://www.lintcode.com/en/problem/find-minimum-in-rotated-sorted-array-ii/)
### Problem
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.
#### Example
Given [4,4,5,6,7,0,1,2] return 0
### 题解
由于此题输入可能有重复元素，因此在`num[mid] == num[end]`时无法使用二分的方法缩小start或者end的取值范围。此时只能使用递增start/递减end逐步缩小范围。
### C++
~~~
class Solution {
public:
/**
* @param num: a rotated sorted array
* @return: the minimum number in the array
*/
int findMin(vector &num) {
if (num.empty()) {
return -1;
}
vector::size_type start = 0;
vector::size_type end = num.size() - 1;
vector::size_type mid;
while (start + 1 < end) {
mid = start + (end - start) / 2;
if (num[mid] > num[end]) {
start = mid;
} else if (num[mid] < num[end]) {
end = mid;
} else {
--end;
}
}
if (num[start] < num[end]) {
return num[start];
} else {
return num[end];
}
}
};
~~~
### Java
~~~
public class Solution {
/**
* @param num: a rotated sorted array
* @return: the minimum number in the array
*/
public int findMin(int[] num) {
if (num == null || num.length == 0) return Integer.MIN_VALUE;
int lb = 0, ub = num.length - 1;
// case1: num[0] < num[num.length - 1]
// if (num[lb] < num[ub]) return num[lb];
// case2: num[0] > num[num.length - 1] or num[0] < num[num.length - 1]
while (lb + 1 < ub) {
int mid = lb + (ub - lb) / 2;
if (num[mid] < num[ub]) {
ub = mid;
} else if (num[mid] > num[ub]){
lb = mid;
} else {
ub--;
}
}
return Math.min(num[lb], num[ub]);
}
}
~~~
### 源码分析
注意`num[mid] > num[ub]`时应递减 ub 或者递增 lb.
### 复杂度分析
最坏情况下 O(n)O(n)O(n), 平均情况下 O(logn)O(\log n)O(logn).
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