LeetCode - Search in Rotated Sorted Array

题目：

Search in Rotated Sorted Array
Total Accepted: 81090 Total Submissions: 277272 Difficulty: Hard

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.

思路：

数组被翻转，我们以中间元素为核心，把它分为两种情况，一种情况是中间元素的左边是有序的，一种是中间元素的右边是有序的。比如{ 0 1 2 4 5 6 7 }被翻转为{ 6 7 0 1 2 4 5 }或{ 2 4 5 6 7 0 1 }。

所以每次判断我们都取轻避重，总是跟有序的这边比较。

package array;

public class SearchInRotatedSortedArray {

    public int search(int[] nums, int target) {
        int len = 0;
        if (nums == null || (len = nums.length) == 0) return 0;
        // { 0 1 2 4 5 6 7 }
        // 1. { 6 7 0 1 2 4 5 }
        // 2. { 2 4 5 6 7 0 1 }
        int l = 0;
        int r = len - 1;
        
        while (l <= r) {
            int mid = (l + r) / 2;
            if (nums[mid] == target) return mid;
            if (nums[mid] < nums[r]) {
                if (target > nums[mid] && target <= nums[r])
                    l = mid + 1;
                else
                    r = mid - 1;
            } else {
                if (nums[l] <= target && target < nums[mid]) 
                    r = mid - 1;
                else
                    l = mid + 1;
            }
        }
        
        return -1;
    }
    
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        int[] nums = { 4, 5, 6, 7, 0, 1, 2 };
        SearchInRotatedSortedArray s = new SearchInRotatedSortedArray();
        System.out.println(s.search(nums, 2));
    }

}