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
难度：medium
题目：假设一个按升序排序的数组在某个未知的轴上旋转。给定一个搜索值，如果能找到则返回其所在数组中的位置，否则返回-1. 假定数组中无重复元素。算法时间复杂度要为 O(log n)
思路：二叉搜索
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;
}
}