[LintCode] Find Minimum 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).

Find the minimum element.

You may assume no duplicate exists in the array.

Example

Given [4, 5, 6, 7, 0, 1, 2] return 0

 

public class Solution {
    /**
     * @param nums: a rotated sorted array
     * @return: the minimum number in the array
     */
    public int findMin(int[] nums) {
        return findMinHelper(nums, 0, nums.length - 1);
    }
    
    private int findMinHelper(int[] nums, int startIdx, int endIdx)
    {
        if(endIdx - startIdx <= 1)
        {
            return nums[startIdx] < nums[endIdx] ? nums[startIdx] : nums[endIdx];
        }
        
        int mid = startIdx + (endIdx - startIdx) / 2;
        //right half not sorted, min value must be in the right half
        //nums[mid] can't be the minimum 
        if(nums[mid] > nums[endIdx])
        {
            return findMinHelper(nums, mid + 1, endIdx);
        }
        //nums[mid] <= nums[endIdx]
        else
        {
            return findMinHelper(nums, startIdx, mid);
        }
    }
}

 

 

 

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