LeetCode 1095. 山脉数组中查找目标值（二分查找）

MountainArray.get(k) - 会返回数组中索引为k 的元素（下标从 0 开始）
MountainArray.length() - 会返回该数组的长度

注意：
对 MountainArray.get 发起超过 100 次调用的提交将被视为错误答案。

示例 1：
输入：array = [1,2,3,4,5,3,1], target = 3
输出：2
解释：3 在数组中出现了两次，下标分别为 2 和 5，我们返回最小的下标 2。

示例 2：
输入：array = [0,1,2,4,2,1], target = 3
输出：-1
解释：3 在数组中没有出现，返回 -1。

提示：
	3 <= mountain_arr.length() <= 10000
	0 <= target <= 10^9
	0 <= mountain_arr.get(index) <= 10^9

/**
 * // This is the MountainArray's API interface.
 * // You should not implement it, or speculate about its implementation
 * class MountainArray {
 *   public:
 *     int get(int index);
 *     int length();
 * };
 */

class Solution {
public:
    int findInMountainArray(int target, MountainArray &mArr) {
        int n = mArr.length(), l = 0, r = n-1, high;
        int lval, rval, mid, midval, mid1, midval1;
        while(l < r)
        {
            mid = l+((r-l)>>1);
            midval = mArr.get(mid);
            midval1 = mArr.get(mid+1);
            if(midval < midval1)
                l = mid+1;
            else
                r = mid;
        }
        high = l;
        l = 0; r = high;
        while(l <= r)
        {
            mid = l+((r-l)>>1);
            midval = mArr.get(mid);
            if(midval == target)
                return mid;
            else if(midval > target)
                r = mid-1;
            else
                l = mid+1;
        }
        l = high+1, r = n-1;
        while(l <= r)
        {
            mid = l+((r-l)>>1);
            midval = mArr.get(mid);
            if(midval == target)
                return mid;
            else if(midval < target)
                r = mid-1;
            else
                l = mid+1;
        }
        return -1;
    }
};