LeetCode--No.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

感覺是一道非常基本的題了，其實就是binary search呀，只不過稍微變形了一下，但是我卻寫了這麼多的bug出來，可見binary search掌握得多不熟練，各種corner case都考慮不到，最後拼拼湊湊，改來改去，終於有了下面這個破破的代碼。

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

所以說對於binary search/recursive的話，最先決定的，一定是終止條件。在這裏就是return了-1或者return了某個值.