【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 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.

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 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: 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 in the array, so we return -1.

Constraints:

1. 3 <= mountain_arr.length() <= 10000
2. 0 <= target <= 10^9
3. 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