*LeetCode-Find Minimum in Rotated Sorted Array

可以o(n)扫一遍 找到不比前一个大的数字 

lgn的方法是 binary search 通过判断head tail mid的大小关系判断pivot在哪一段

public class Solution {
    public int findMin(int[] nums) {
        int head = 0;
        int tail = nums.length - 1;
        while ( head < tail ){
            int mid = head + ( tail - head ) /2;
            if ( nums [ head ] > nums [ mid ] )
                tail = mid;
            else if ( nums [ mid ] > nums [ tail ] )
                head = mid + 1;
            else
                return nums [ head ];
        }
        return nums [ head ];
    }
}