Leetcode: Search in Rotated Sorted Array

Suppose a sorted array 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.

根据binary search，每次我们都可以切掉一半的数据，所以算法的时间复杂度是O(logn)，空间复杂度是O(1)。

class Solution {
    public int search(int[] nums, int target) {
        if (nums == null || nums.length == 0) return -1;
        int l = 0, r = nums.length - 1;
        while (l <= r) {
            int m = l + (r - l) / 2;
            if (nums[m] == target) return m;
            if (nums[m] >= nums[l]) {
                // [l, m] is in order
                if (nums[l] <= target && target < nums[m]) {
                    r = m - 1;
                }
                else l = m + 1;
            }
            else {
                // [m, r] is in order
                if (nums[m] < target && target <= nums[r]) {
                    l = m + 1;
                }
                else r = m - 1;
            }
        }
        return -1;
    }
}