算法精粹(algorithm-essentials)

感谢soulmachine@github提供内容
## Search in Rotated Sorted Array


### 描述

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`).

You are given a target value to search. If found in the array return its index, otherwise return `-1`.

You may assume no duplicate exists in the array.


### 分析

一个有序数组被循环右移,只可能有一下两种情况:

```
7 │
6 │
─────┼───────────
│ 5
│ 4
│ 3
│ 2
│ 1
```

```
7 │
6 │
5 │
4 │
3 │
───────────┼───────────
│ 2
│ 1
```

本题依旧可以用二分查找,难度主要在于左右边界的确定。仔细观察上面两幅图,我们可以得出如下结论:

如果`A[left] <= A[mid]`,那么`[left,mid]` 一定为单调递增序列。


### 代码

{% if book.java %}
```java
// Search in Rotated Sorted Array
// Time Complexity: O(log n),Space Complexity: O(1)
public class Solution {
public int search(int[] nums, int target) {
int first = 0, last = nums.length;
while (first != last) {
final int mid = first + (last - first) / 2;
if (nums[mid] == target)
return mid;
if (nums[first] <= nums[mid]) {
if (nums[first] <= target && target < nums[mid])
last = mid;
else
first = mid + 1;
} else {
if (nums[mid] < target && target <= nums[last-1])
first = mid + 1;
else
last = mid;
}
}
return -1;
}
};
```
{% endif %}


{% if book.cpp %}
```cpp
// Search in Rotated Sorted Array
// Time Complexity: O(log n),Space Complexity: O(1)
class Solution {
public:
int search(const vector& nums, int target) {
int first = 0, last = nums.size();
while (first != last) {
const int mid = first + (last - first) / 2;
if (nums[mid] == target)
return mid;
if (nums[first] <= nums[mid]) {
if (nums[first] <= target && target < nums[mid])
last = mid;
else
first = mid + 1;
} else {
if (nums[mid] < target && target <= nums[last-1])
first = mid + 1;
else
last = mid;
}
}
return -1;
}
};
```
{% endif %}


### 相关题目

* [Search in Rotated Sorted Array II](search-in-rotated-sorted-array-ii.md)
* [Find Minimum in Rotated Sorted Array](find-minimum-in-rotated-sorted-array.md)
* [Find Minimum in Rotated Sorted Array II](find-minimum-in-rotated-sorted-array-ii.md)