LeetCode Next Greater Element I

寻找下一个更大元素

最新推荐文章于 2020-06-14 02:02:27 发布

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本文介绍了一个算法问题：给定两个不包含重复元素的数组nums1和nums2，其中nums1的元素是nums2的子集。任务是找到nums1中每个元素在nums2中对应位置的下一个更大的数。如果不存在，则输出-1。

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.
For number 4 in the first array, there is no next greater number for it in the second array, so output -1.
Note:
All elements in nums1 and nums2 are unique.
The length of both nums1 and nums2 would not exceed 1000.

class Solution(object):
    def nextGreaterElement(self, findNums, nums):
        """
        :type findNums: list[int]
        :type nums: list[int]
        :rtype: list[int]
        """
        res = []
        for i in findNums:
            j = nums.index(i)
            flag=1
            for k in nums[j:]:
                if k > i:
                    res.append(k)
                    flag=0
                    break
            if flag==1:
                res.append(-1)
        return  res