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

O(logn)

end = mid;

start = mid 而不是！ start = nums[mid]  !!!

class Solution {
    public int search(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return -1;
        }
        int start = 0;
        int end = nums.length - 1;
        int pivot = nums[0];
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (nums[mid] == target) {
                return mid;
            }else if (nums[mid] >= pivot) {
                if (target >= nums[start] && target <= nums[mid]) {
                    end = mid;
                } else {
                    start = mid;
                }
            } else {
                if (target <= nums[end] && target >= nums[mid]) {
                    start  = mid;
                }else {
                    end = mid;
                }
            }
        }
        if (nums[start] == target) {
            return start;
        }else if (nums[end] == target){
            return end;
        }else {
            return -1;
        }
    }
}