[Java]LeetCode1095. 山脉数组中查找目标值 | Find in Mountain Array

时间 2019-11-11

标签 java leetcode1095 leetcode 山脉数组查找目标 mountain array 栏目 Java 繁體版

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(This problem is an interactive problem.)git

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

A.length >= 3
There exists some i with 0 < i < A.length - 1 such that:
- A[0] < A[1] < ... A[i-1] < A[i]
- A[i] > A[i+1] > ... > A[A.length - 1]

Given a mountain array mountainArr, return the minimum index such that mountainArr.get(index) == target. If such an index doesn't exist, return -1.数组

You can't access the mountain array directly. You may only access the array using a MountainArray interface:微信

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.spa

Example 1:code

Input: array = [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:htm

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

Constraints:blog

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

（这是一个交互式问题）索引

给你一个山脉数组 mountainArr，请你返回可以使得 mountainArr.get(index) 等于 target 最小的下标 index 值。

若是不存在这样的下标 index，就请返回 -1。

所谓山脉数组，即数组 A 假如是一个山脉数组的话，须要知足以下条件：

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，必须经过 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致使比赛资格被取消。

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

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，咱们返回最小的下标 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

Runtime: 0 ms

C++:

 1 /**
 2  * // This is the MountainArray's API interface.
 3  * // You should not implement it, or speculate about its implementation
 4  * class MountainArray {
 5  *   public:
 6  *     int get(int index);
 7  *     int length();
 8  * };
 9  */
10 class Solution {
11 public:
12     int findInMountainArray(int target, MountainArray &mountainArr) {
13         int n = mountainArr.length();
14         int low = 1, high = n - 1;
15         while (low < high) {
16             int mid = (low + high) / 2;
17 
18             int a = mountainArr.get(mid - 1), b = mountainArr.get(mid), c = mountainArr.get(mid + 1);
19 
20             if (a < b && b > c) {
21                 low = high = mid;
22                 break;
23             }
24 
25             if (a < b && b < c)
26                 low = mid + 1;
27             else if (a > b && b > c)
28                 high = mid - 1;
29             else
30                 assert(false);
31         }
32 
33         int mountain = low;
34 
35         low = 0;
36         high = mountain;
37 
38         while (low < high) {
39             int mid = (low + high) / 2;
40 
41             if (mountainArr.get(mid) >= target)
42                 high = mid;
43             else
44                 low = mid + 1;
45         }
46 
47         if (mountainArr.get(low) == target)
48             return low;
49 
50         low = mountain;
51         high = n - 1;
52 
53         while (low < high) {
54             int mid = (low + high) / 2;
55 
56             if (mountainArr.get(mid) <= target)
57                 high = mid;
58             else
59                 low = mid + 1;
60         }
61 
62         if (mountainArr.get(low) == target)
63             return low;
64 
65         return -1;
66     }
67 };

0ms

Java:

 1 /**
 2  * // This is MountainArray's API interface.
 3  * // You should not implement it, or speculate about its implementation
 4  * interface MountainArray {
 5  *     public int get(int index) {}
 6  *     public int length() {}
 7  * }
 8  */
 9  
10 class Solution {
11        int findInMountainArray(int target, MountainArray A) {
12         int n = A.length(), l, r, m, peak = 0;
13         // find index of peak
14         l  = 0;
15         r = n - 1;
16         while (l < r) {
17             m = (l + r) / 2;
18             if (A.get(m) < A.get(m + 1))
19                 l = peak = m + 1;
20             else
21                 r = m;
22         }
23         // find target in the left of peak
24         l = 0;
25         r = peak;
26         while (l <= r) {
27             m = (l + r) / 2;
28             if (A.get(m) < target)
29                 l = m + 1;
30             else if (A.get(m) > target)
31                 r = m - 1;
32             else
33                 return m;
34         }
35         // find target in the right of peak
36         l = peak;
37         r = n - 1;
38         while (l <= r) {
39             m = (l + r) / 2;
40             if (A.get(m) > target)
41                 l = m + 1;
42             else if (A.get(m) < target)
43                 r = m - 1;
44             else
45                 return m;
46         }
47         return -1;
48     }
49 }

24ms

Python3:

 1 # """
 2 # This is MountainArray's API interface.
 3 # You should not implement it, or speculate about its implementation
 4 # """
 5 #class MountainArray:
 6 #    def get(self, index: int) -> int:
 7 #    def length(self) -> int:
 8 
 9 class Solution:
10     def findInMountainArray(self, target: int, mo: 'MountainArray') -> int:
11         mlen = mo.length()
12         left, right, peak = 0, mlen - 1, 0
13         while left < right:
14             mid = left + (right - left) // 2
15             if mo.get(mid + 1) > mo.get(mid):
16                 peak = mid + 1
17                 left = mid + 1
18             else:
19                 right = mid - 1
20         
21         left, right = 0, peak
22         while left <= right:
23             mid = left + (right - left) // 2
24             val = mo.get(mid)
25             if val < target:
26                 left = mid + 1
27             elif val == target:
28                 return mid
29             else:
30                 right = mid - 1
31 
32         left, right = peak, mlen - 1
33         while left <= right:
34             mid = left + (right - left) // 2
35             val = mo.get(mid)
36             if val < target:
37                 right = mid - 1
38             elif val == target:
39                 return mid
40             else:
41                 left = mid + 1
42                 
43         return -1