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33. Search in Rotated Sorted Array
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
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2018-09-16 17:45
*/
public static int search(int[] nums, int target) {
int result = -1;
if (null == nums || nums.length == 0) {
return result;
}
int firstNum = nums[0];
int lastNum = nums[nums.length - 1];
int separator = -1;
if (firstNum > lastNum) {
int head = 0;
int tail = nums.length - 1;
while (head <= tail) {
int mid = head + (tail - head) / 2;
int midNum = nums[mid];
if (midNum > nums[mid + 1]) {
separator = mid;
break;
}
if (midNum >= firstNum) {
head = mid + 1;
}
if (midNum < lastNum) {
tail = mid - 1;
}
}
}
if (separator == -1) {
return binarySearch(nums, target, 0, nums.length - 1);
} else {
if (firstNum <= target && target <= nums[separator]) {
return binarySearch(nums, target, 0, separator);
} else {
return binarySearch(nums, target, separator + 1, nums.length - 1);
}
}
}
private static int binarySearch(int[] nums, int target, int headIndex, int tailIndex) {
int head = headIndex;
int tail = tailIndex;
while (head <= tail) {
int mid = head + (tail - head) / 2;
int midNum = nums[mid];
if (midNum == target) {
return mid;
}
if (target <= midNum) {
tail = mid - 1;
}
if (midNum < target) {
head = mid + 1;
}
}
return -1;
}