Maximum Number in Mountain Sequence

01/18/2018 update
Solution:

public class Solution {
    /**
     * @param nums a mountain sequence which increase firstly and then decrease
     * @return then mountain top
     */
    public int mountainSequence(int[] nums) {
        // Write your code here
        if (nums == null || nums.length == 0) {
            return 0;
        }
        int start = 0;
        int end = nums.length - 1;
        while (start + 1 < end) {
            int mid = (start + end) / 2;
            if (nums[mid] > nums[mid + 1]) {
                end = mid;
            } else {
                start = mid;
            }
        }
        return Math.max(nums[start], nums[end]);
    }
}

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思路：
虽然写出来通过了，但是很复杂。九章的答案很简洁。直接比较nums[mid]>nums[mid+1]，如果成立则mountain在左边，否则mountain在右边。最后取nums[start] nums[end]最大的一个就好了。循环条件写成start+1 < end的好处是，最后退出的时候可以再检查一遍start和end，确保不会漏掉，而且不会有死循环。