Find Minimum in Rotated Sorted Array
最后更新于：2022-04-02 01:09:10
# Find Minimum in Rotated Sorted Array
### Source
- leetcode: [Find Minimum in Rotated Sorted Array | LeetCode OJ](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/)
- lintcode: [(159) Find Minimum in Rotated Sorted Array](http://www.lintcode.com/en/problem/find-minimum-in-rotated-sorted-array/)
### 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.
#### Example
Given `[4, 5, 6, 7, 0, 1, 2]` return `0`
#### Note
You may assume no duplicate exists in the array.
### 题解
如前节所述，对于旋转数组的分析可使用画图的方法，如下图所示，升序数组经旋转后可能为如下两种形式。
最小值可能在上图中的两种位置出现，如果仍然使用数组首部元素作为target去比较，则需要考虑图中右侧情况。**使用逆向思维分析，如果使用数组尾部元素分析，则无需图中右侧的特殊情况。**不过考虑在内的话也算是一种优化。
### 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]) {
end = mid;
} else {
start = mid;
}
}
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 {
lb = mid;
}
}
return Math.min(num[lb], num[ub]);
}
}
~~~
### 源码分析
仅需注意使用`num[end]`(使用 num[lb]不是那么直观)作为判断依据即可，由于题中已给无重复数组的条件，故无需处理`num[mid] == num[end]`特殊条件。
### 复杂度分析
由于无重复元素，平均情况下复杂度为 O(logn)O(\log n)O(logn).
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