[LeetCode] 1095. Find in Mountain Array

(This problem is an interactive problem.)

You may recall that an array arr is a mountain array if and only if:

arr.length >= 3
There exists some i with 0 < i < arr.length - 1 such that:
arr[0] < arr[1] < … < arr[i - 1] < arr[i]
arr[i] > arr[i + 1] > … > arr[arr.length - 1]
Given a mountain array mountainArr, return the minimum index such that mountainArr.get(index) == target. If such an index does not exist, return -1.

You cannot access the mountain array directly. You may only access the array using a MountainArray interface:

MountainArray.get(k) returns the element of the array at index k (0-indexed).
MountainArray.length() returns the length of the array.
Submissions making more than 100 calls to MountainArray.get will be judged Wrong Answer. Also, any solutions that attempt to circumvent the judge will result in disqualification.

Example 1:
Input: mountainArr = [1,2,3,4,5,3,1], target = 3
Output: 2
Explanation: 3 exists in the array, at index=2 and index=5. Return the minimum index, which is 2.

Example 2:
Input: mountainArr = [0,1,2,4,2,1], target = 3
Output: -1
Explanation: 3 does not exist in the array, so we return -1.

Constraints:
3 <= mountainArr.length() <= 104
0 <= target <= 109
0 <= mountainArr.get(index) <= 109

山脉数组中查找目标值。

（这是一个 交互式问题 ）

给你一个 山脉数组 mountainArr，请你返回能够使得 mountainArr.get(index) 等于 target 最小 的下标 index 值。

如果不存在这样的下标 index，就请返回 -1。

何为山脉数组？如果数组 A 是一个山脉数组的话，那它满足如下条件：

首先，A.length >= 3

其次，在 0 < i < A.length - 1 条件下，存在 i 使得：

A[0] < A[1] < … A[i-1] < A[i]
A[i] > A[i+1] > … > A[A.length - 1]

你将 不能直接访问该山脉数组，必须通过 MountainArray 接口来获取数据：

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。此外，任何试图规避判题系统的解决方案都将会导致比赛资格被取消。

为了帮助大家更好地理解交互式问题，我们准备了一个样例 “答案”：https://leetcode-cn.com/playground/RKhe3ave，请注意这 不是一个正确答案。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/find-in-mountain-array
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路

做这个题如果对山脉数组的定义不是很明确，可以先做一下852题。这个题可以跟162，852一起做。山脉数组就是只有一个峰值 peak，峰值左边的元素是单调递增的，右边的元素是单调递减的。本题不允许你直接访问 input 数组，你只能通过length()函数知道数组的长度，通过get()函数得知某一个数字的值。只能用二分法做。

这个题的思路会用到三次二分法，第一次帮助找到 peak 元素，第二次去找 peak 元素左半边是否存在 target，第三次在 peak 元素右半边查找 target。注意体会不同模板比较适合的场景。

复杂度

时间O(logn)
空间O(1)

代码

Java实现

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

class Solution {
    public int findInMountainArray(int target, MountainArray mountainArr) {
        int n = mountainArr.length();
        int peak = 0;
        int mid = 0;
        // find index of peak
        // 因为mid + 1可能会有越界的情况所以这里给了闭区间
        // 下标都在index范围内
        int left = 0;
        int right = n - 1;
        while (left < right) {
            mid = left + (right - left) / 2;
            if (mountainArr.get(mid) < mountainArr.get(mid + 1)) {
                left = peak = mid + 1;
            } else {
                right = mid;
            }
        }

        /*
        找peak元素部分的二分也可以这样写，需要仔细体会
        int left = 0;
        int right = n - 1;
        while (left + 1 < right) {
            mid = left + (right - left) / 2;
            if (mountainArr.get(mid) < mountainArr.get(mid + 1)) {
                left = mid;
            } else {
                right = mid;
            }
        }
        if (mountainArr.get(left) > mountainArr.get(right)) {
            peak = left;
        } else {
            peak = right;
        }
        */

        // find target in the left of peak
        left = 0;
        right = peak;
        while (left <= right) {
            mid = left + (right - left) / 2;
            if (mountainArr.get(mid) < target) {
                left = mid + 1;
            } else if (mountainArr.get(mid) > target) {
                right = mid - 1;
            } else {
                return mid;
            }
        }

        // find target in the right of peak
        left = peak;
        right = n - 1;
        while (left <= right) {
            mid = left + (right - left) / 2;
            if (mountainArr.get(mid) > target) {
                left = mid + 1;
            } else if (mountainArr.get(mid) < target) {
                right = mid - 1;
            } else {
                return mid;
            }
        }
        return -1;
    }
}

相关题目

162. Find Peak Element
852. Peak Index in a Mountain Array
1095. Find in Mountain Array