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# 33.Search-in-Rotated-Sorted-Array

## 题目描述

Suppose an array sorted in ascending order 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.
Your algorithm's runtime complexity must be in the order of O(log n).
Example 1:
Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4
Example 2:
Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

## 代码

Complexity Analysis
• Time complexity: O(logN).
• Space complexity: O(1).
public class Solution {
public int search(int[] A, int target) {
if (A == null || A.length == 0) return -1;
int start = 0;
int end = A.length - 1;
int mid;
while (start + 1 < end) {
mid = start + (end - start) / 2;
if (A[mid] == target) {
return mid;
}
if (A[start] < A[mid]) {
if (A[start] <= target && target <= A[mid]) {
end = mid;
} else {
start = mid;
}
} else {
if (A[mid] <= target && target <= A[end]) {
start = mid;
} else {
end = mid;
}
}
}
if (A[start] == target) {
return start;
}
if (A[end] == target) {
return end;
}
return -1;
}
}
class Solution {
public int search(int[] nums, int target) {
int start = 0, end = nums.length - 1;
while (start <= end) {
int mid = start + (end - start) / 2;
if (nums[mid] == target) return mid;
else if (nums[mid] >= nums[start]) {
if (target >= nums[start] && target < nums[mid]) end = mid - 1;
else start = mid + 1;
}
else {
if (target <= nums[end] && target > nums[mid]) start = mid + 1;
else end = mid - 1;
}
}
return -1;
}
}