【leetcode】1095. Find in Mountain Array
题目如下:
(This problem is an interactive problem.)
You may recall that an array
A
is a mountain array if and only if:
A.length >= 3
- There exists some
i
with0 < 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 minimumindex
such thatmountainArr.get(index) == target
. If such anindex
doesn't exist, return-1
.You can't 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 indexk
(0-indexed).MountainArray.length()
returns the length of the array.Submissions making more than
100
calls toMountainArray.get
will be judged Wrong Answer. Also, any solutions that attempt to circumvent the judge will result in disqualification.
Example 1:
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:
Input: array = [0,1,2,4,2,1], target = 3 Output: -1 Explanation: 3 does not exist inthe array,
so we return -1.
Constraints:
3 <= mountain_arr.length() <= 10000
0 <= target <= 10^9
0 <= mountain_arr.get(index) <= 10^9
解题思路:我的解法是二分查找。mountain array 数组的特点是有一个顶点,顶点左边的区间是单调递增,右边的区间是单调递减。所以首先是找出顶点的下标,对于任意一个点mid,如果值比(mid-1)和(mid+1)都大,表示这个是顶点;如果mid的值大于(mid+1),表示mid处于下降区间,令high = mid - 1;如果mid的值大于(mid-1),表示mid处于上升区间,令low = mid + 1;最终可以计算出顶点top。接下来再对左边的上升区间做二分查找求target,如果找到则返回对应小标;没有的话继续对右边的下降区间用二分查找。
代码如下:
# """ # This is MountainArray's API interface. # You should not implement it, or speculate about its implementation # """ #class MountainArray(object): # def get(self, index): # """ # :type index: int # :rtype int # """ # # def length(self): # """ # :rtype int # """ class Solution(object): def findInMountainArray(self, target, mountain_arr): """ :type target: integer :type mountain_arr: MountainArray :rtype: integer """ length = mountain_arr.length() low = 0 high = length - 1 while low <= high: mid = (low + high)/2 mid_val = mountain_arr.get(mid) mid_l_val,mid_h_val = -float('inf'),-float('inf') if mid - 1 >= 0: mid_l_val = mountain_arr.get(mid-1) if mid + 1 <= high: mid_h_val = mountain_arr.get(mid+1) if mid_val > mid_l_val and mid_val > mid_h_val: break elif mid_val > mid_h_val: high = mid - 1 elif mid_val > mid_l_val: low = mid + 1 #left low,high = 0,mid res = -1 while low <= high: mid = (low + high)/2 mid_val = mountain_arr.get(mid) if target == mid_val: res = mid break elif target > mid_val: low = mid + 1 else: high = mid - 1 if res != -1:return res low, high = mid,length-1 while low <= high: mid = (low + high)/2 mid_val = mountain_arr.get(mid) if target == mid_val: res = mid break elif target < mid_val: low = mid + 1 else: high = mid - 1 return res