[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.

Example

Given [4, 5, 6, 7, 0, 1, 2] return 0

直接二分法查找。最终会有两种情况，即指针走到相邻位置，a,b中a大于b或者b大于a。a大于b时，取最左侧。反之，用if(nums[mid] >= nums[left]) left=mid+1跳过a。

public class Solution {
    /**
     * @param nums: a rotated sorted array
     * @return: the minimum number in the array
     */
    public int findMin(int[] nums) {
        return helper(nums, 0, nums.length - 1);
    }
    
    int helper(int[] nums, int left, int right) {
        
        while(left < right) {
            if(nums[left] < nums[right]) return nums[left];

            int mid = (left + right) / 2;
            if(nums[mid] >= nums[left]) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        
        return nums[left];
    }
}