【LeetCode】496. Next Greater Element I【E】【94】

最新推荐文章于 2021-10-26 01:11:44 发布

原创最新推荐文章于 2021-10-26 01:11:44 发布 · 747 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode题解 #栈

Python 同时被 3 个专栏收录

176 篇文章

订阅专栏

leetcode

175 篇文章

订阅专栏

算法

157 篇文章

订阅专栏

本文介绍了一种使用栈高效解决寻找数组中下一个更大元素的问题。通过观察递减序列及其后第一个较大数的特点，利用栈保持递减子序列，并巧妙地找到对应每个元素的下一个更大数。

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.

Subscribe to see which companies asked this question.

o（n^2）的算法还是很好想的是被注释掉的那部分

看见题目标签说stack

然后看了给的solution，很精致的用stack的方法

思路如下

Key observation:
Suppose we have a decreasing sequence followed by a greater number
For example [5, 4, 3, 2, 1, 6] then the greater number 6 is the next greater element for all previous numbers in the sequence

We use a stack to keep a decreasing sub-sequence, whenever we see a number x greater than stack.peek() we pop all elements less than x and for all the popped ones, their next greater element is x
For example [9, 8, 7, 3, 2, 1, 6]

The stack will first contain [9, 8, 7, 3, 2, 1] and then we see 6 which is greater than 1 so we pop 1 2 3 whose next greater element should be 6

class Solution(object):
    def nextGreaterElement(self, findNums, nums):
        a = findNums
        b = nums 

        m = {}

        s = []

        for i in b:
            while len(s) and s[-1] < i:
                m[s.pop()] = i
            s += i,
        res = []
        for i in a:
            res += m.get(i,-1),

        return res


        """

        #a.sort()
        #b.sort()

        #print a,b

        res = []
        for i in a:
            pos = b.index(i)
            #bb = b[pos:]

            j = pos
            while j < len(b):
                if b[j] > i:
                    res += b[j],
                    break
                j += 1
            if j == len(b):
                res += -1,

        return res

        :type findNums: List[int]
        :type nums: List[int]
        :rtype: List[int]
        """