Leetcode-Algorithms Next Greater Element I

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本文介绍了一种算法，用于解决在两个数组中寻找Next Greater Element的问题。即，在nums1中的每个元素在nums2中找到第一个大于它的元素，若不存在则返回-1。文中提供了一个简单直接的实现方法，并给出了优化后的O(n)解法。

You are given two arrays (without duplicates) nums1 and nums2 where nums1’s elements are subset of nums2. Find all the next greater numbers for nums1’s elements in the corresponding places of nums2.

The Next Greater Number of a number x in nums1 is the first greater number to its right in nums2. If it does not exist, output -1 for this number.

Example 1:
Input: nums1 = [4,1,2], nums2 = [1,3,4,2].
Output: [-1,3,-1]
Explanation:
For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1.
For number 1 in the first array, the next greater number for it in the second array is 3.
For number 2 in the first array, there is no next greater number for it in the second array, so output -1.

Example 2:
Input: nums1 = [2,4], nums2 = [1,2,3,4].
Output: [3,-1]
Explanation:
For number 2 in the first array, the next greater number for it in the second array is 3.

Note:
All elements in nums1 and nums2 are unique.
The length of both nums1 and nums2 would not exceed 1000.

给出两个不含重复元素的array nums1, nums2。 nums1 是nums2的子集（nums2包含nunms1的所有元素）。

Next Greater Element 指的是找出在nums1里面的每个元素在nums2中相同元素的右边比它本身大的元素，如果没有或者比其小则为-1。

class Solution(object):
    def nextGreaterElement(self, findNums, nums):
        """
        :type findNums: List[int]
        :type nums: List[int]
        :rtype: List[int]
        """
        result = []

        for c in findNums:
            pos = nums.index(c)
            while pos <= (len(nums) - 1):
                if pos == len(nums) - 1:
                    result.append(-1)
                    break
                else:
                    if nums[pos+1] > c:
                        result.append(nums[pos+1])
                        break
                    else:
                        pos += 1
        return result

附上leeetcode答案 O(n)：

d = {}
        st = []
        ans = []

        for x in nums:
            while len(st) and st[-1] < x:
                d[st.pop()] = x
            st.append(x)

        for x in findNums:
            ans.append(d.get(x, -1))

        return ans