重复的解决方法是stack记录下标,而不是记录数据,每次有了next,直接存到res的相应位置 最后的res就相当于一个哈希表 publicint[] nextGreaterElements2(int[] nums) {/*相对第一题的改变时数组成了循环数组,遍历到最后一个数之后可以再从第一个数开始 自己想的方案是暴力解,两层for 看了答案,这种循环数组
这个题目就是用stack, 分别将nums2里面每个对应的next bigger number找到, 如果没有, 那么为-1, 并且存到d里面. T :O(m) m = len(nums) Code classSolution(object):defnextGreaterElement(self, findNums, nums): stack,d, ans=[],{}, []fornuminnums:whilestackandnum > stack[-1]: d[stack.p...
来自专栏 · LeetCode刷题记录 1 人赞同了该文章 503. Next Greater Element II 难度:m class Solution: def nextGreaterElements(self, nums: List[int]) -> List[int]: if not nums: return [] stack = [] res = [-1]*len(nums) for i in range(len(nums)): while stack and nums[stack[...
Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element. The Next Greater Number of a number x is the first greater number to its traversing-order next in the array, which means you could search ci...
Next Greater Element I? Leetcode 496. Next Greater Element I的时间复杂度是多少? 1. Description 2. Solution **解析:**Version 1,由于元素是唯一的,通过循环找出每个nums2中的满足条件结果保存到字典中,遍历nums1,获得结果。Version 2通过使用栈来寻找满足条件的结果,减少搜索时间。 Version 1 代码语言:...
Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element. The Next Greater Number of a number x is the first greater number to its traversing-order next in the array, which means you could search ci...
## LeetCode 496fromtypingimportListclassSolution:defnextGreaterElement(self,nums1:List[int],nums2:List[int])->List[int]:res={}stack=[]foriinnums2[::-1]:## 从 num2 最右边的元素开始遍历whilestackandi>=stack[-1]:## 如果栈非空,且该元素比栈顶(列表最右边)元素大stack.pop()## 清空这个...
LeetCode题目:503. Next Greater Element II Given acircular array (the next element of the last element is the first element of the array),print the Next Greater Number for every element. The Next Greater Number of a number x is the first greater number to its traversing-order next in the...
Next Greater Element II 2. Solution 解析:Version 1,由于元素不是唯一的,需要循环查找,因此先将nums复制一遍,通过循环每次都查找当前元素之后的n-1位数字。Version 2通过使用栈来寻找满足条件的结果,栈中保持是数字的索引位置,由于需要循环查找,因此需要查找两次nums,并且第二次查找不需要保持数字索引。
[LeetCode] 496. Next Greater Element I 题目内容 https://leetcode-cn.com/problems/next-greater-element-i/ 给定两个没有重复元素的数组 nums1 和 nums2 ,其中nums1 是 nums2 的子集。找到 nums1 中每个元素在 nums2 中的下一个比其大的值。 ......