33. Search in Rotated Sorted Array - jiejackyzhang/leetcode-note GitHub Wiki

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.

比较straightforward的方法是先找出min，然后分成两部分进行binary search。

注意到如果取中间元素，至少有一边是sorted的。判断target是否在sorted的一边，否则在另一边，相应地update左右指针，继续上述过程。

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