LeetCode 496. Next Greater Element I 单调栈

The next greater element of some element x in an array is the first greater element that is to the right of x in the same array.

You are given two distinct 0-indexed integer arrays nums1 and nums2, where nums1 is a subset of nums2.

For each \(0 <= i < nums1.length\), find the index \(j\) such that \(nums1[i] == nums2[j]\) and determine the next greater element of \(nums2[j]\) in nums2. If there is no next greater element, then the answer for this query is -1.

Return an array ans of length nums1.length such that ans[i] is the next greater element as described above.

Solution

利用单调栈 \(stack\), 如果栈顶元素 \(stack[-1]<nums2[i]\)，说明此时可以存在 \(stack[-1]\rightarrow nums2[i]\) 的映射，直到这个条件不满足。所以最后存留在 \(stack\) 里面的都是无法配对的，此时赋予 \(-1\)

点击查看代码

class Solution(object):
    def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
        l2 = len(nums2)
        l1 = len(nums1)

        mp = {}
        ans = []
        stack = []
        
        for i in range(l2):
            while stack and stack[-1]<nums2[i]:
                mp[stack[-1]] = nums2[i]
                stack.pop()
            stack.append(nums2[i])
        
        for ele in stack:
            mp[ele] = -1
        
        for i in range(l1):
            ans.append(mp[nums1[i]])
        
        return ans