33. Search in Rotated Sorted Array

摘要： 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])....

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

難度：medium

題目：假設一個按升序排序的陣列在某個未知的軸上旋轉。

給定一個搜尋值，如果能找到則返回其所在陣列中的位置，否則返回－1. 假定陣列中無重複元素。演算法時間複雜度要為O(log n)

思路：二叉搜尋

Runtime: 10 ms, faster than 21.83% of Java online submissions for Search in Rotated Sorted Array.

Memory Usage: 26.8 MB, less than 41.44% of Java online submissions for Search in Rotated Sorted Array.

class Solution {
public int search(int[] nums, int target) {
int left = 0; 
int right = nums.length - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (target == nums[mid]) {
return mid;
}

// left < mid </> right
if (nums[left] < nums[mid]) {
if (target > nums[mid] || target < nums[left]) {
left = mid + 1;
} else {
right = mid - 1;
}
} else if (nums[left] > nums[mid]) { // left > mid </> right
if (target < nums[mid] || target > nums[right]) {
right = mid - 1;
} else {
left = mid + 1;
}
} else { // left = mid
left += 1;
}
}

return -1;
}
}