1095. 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/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，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() <= 100000 <= target <= 10^90 <= mountain_arr.get(index) <= 10^9原站题解
cpp 解法, 执行用时: 0 ms, 内存消耗: 7.1 MB, 提交时间: 2023-09-27 09:50:54
/**
* // 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 {
int binary_search(MountainArray &mountain, int target, int l, int r, int key(int)) {
target = key(target);
while (l <= r) {
int mid = (l + r) / 2;
int cur = key(mountain.get(mid));
if (cur == target) {
return mid;
} else if (cur < target) {
l = mid + 1;
} else {
r = mid - 1;
}
}
return -1;
}
public:
int findInMountainArray(int target, MountainArray &mountainArr) {
int l = 0, r = mountainArr.length() - 1;
while (l < r) {
int mid = (l + r) / 2;
if (mountainArr.get(mid) < mountainArr.get(mid + 1)) {
l = mid + 1;
} else {
r = mid;
}
}
int peak = l;
int index = binary_search(mountainArr, target, 0, peak, [](int x) -> int{return x;});
if (index != -1) {
return index;
}
return binary_search(mountainArr, target, peak + 1, mountainArr.length() - 1, [](int x) -> int{return -x;});
}
};
java 解法, 执行用时: 0 ms, 内存消耗: 42 MB, 提交时间: 2023-09-27 09:50:06
/**
* // 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 l = 0, r = mountainArr.length() - 1;
while (l < r) {
int mid = (l + r) / 2;
if (mountainArr.get(mid) < mountainArr.get(mid + 1)) {
l = mid + 1;
} else {
r = mid;
}
}
int peak = l;
int index = binarySearch(mountainArr, target, 0, peak, true);
if (index != -1) {
return index;
}
return binarySearch(mountainArr, target, peak + 1, mountainArr.length() - 1, false);
}
public int binarySearch(MountainArray mountainArr, int target, int l, int r, boolean flag) {
if (!flag) {
target *= -1;
}
while (l <= r) {
int mid = (l + r) / 2;
int cur = mountainArr.get(mid) * (flag ? 1 : -1);
if (cur == target) {
return mid;
} else if (cur < target) {
l = mid + 1;
} else {
r = mid - 1;
}
}
return -1;
}
}
python3 解法, 执行用时: 52 ms, 内存消耗: 16.6 MB, 提交时间: 2023-09-27 09:49:47
# """
# This is MountainArray's API interface.
# You should not implement it, or speculate about its implementation
# """
#class MountainArray:
# def get(self, index: int) -> int:
# def length(self) -> int:
def binary_search(mountain, target, l, r, key=lambda x: x):
target = key(target)
while l <= r:
mid = (l + r) // 2
cur = key(mountain.get(mid))
if cur == target:
return mid
elif cur < target:
l = mid + 1
else:
r = mid - 1
return -1
class Solution:
def findInMountainArray(self, target: int, mountain_arr: 'MountainArray') -> int:
l, r = 0, mountain_arr.length() - 1
while l < r:
mid = (l + r) // 2
if mountain_arr.get(mid) < mountain_arr.get(mid + 1):
l = mid + 1
else:
r = mid
peak = l
index = binary_search(mountain_arr, target, 0, peak)
if index != -1:
return index
index = binary_search(mountain_arr, target, peak + 1, mountain_arr.length() - 1, lambda x: -x)
return index
golang 解法, 执行用时: 4 ms, 内存消耗: 3.1 MB, 提交时间: 2023-09-27 09:47:18
/**
* // This is the MountainArray's API interface.
* // You should not implement it, or speculate about its implementation
* type MountainArray struct {
* }
*
* func (this *MountainArray) get(index int) int {}
* func (this *MountainArray) length() int {}
*/
// 内置sort.Search 二分搜索
func findInMountainArray(target int, arr *MountainArray) int {
n := arr.length()
peak := sort.Search(n-1, func(i int) bool { return arr.get(i) >= arr.get(i+1) })
i := sort.Search(peak, func(i int) bool { return arr.get(i) >= target })
if arr.get(i) != target {
i = peak + sort.Search(n-1-peak, func(i int) bool { return arr.get(peak+i) <= target })
if arr.get(i) != target {
return -1
}
}
return i
}