想到了用哈希表存这个数的位置,但是没有想到可以直接用哈希表存next great,用栈存还没找到的数,没遍历一个数就考察栈中的元素小,小的话,这个数就是栈中数的next great,栈中的数肯定是下大上小。 publicint[] nextGreaterElement(int[] nums1,int[] nums2) {/*通过map建立当前元素和其next great的映射 ...
Next Greater Element II 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 m...
3 解题代码 ## 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()...
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...
题目地址:https://leetcode.com/problems/minesweeper/description/ 题目描述 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 great...
若栈非空,则此时栈顶数即为cur对应位置后最近较大数 将以上结果存入字典hashTable中,便于生成结果 nums2遍历结束后,遍历nums1 以当前遍历数,向hashTable中取值,并存入结果列表res 最后返回res 代码如下: class Solution(object): def nextGreaterElement(self,nums1,nums2): ...
question 解法: 小伙伴的思路更加简单,时间复杂度O(n),leetcode上击败98% 思路,还是使用stack,让我们来走一遍例子 dic= {} s = [] [1,3,4,2] s = [1] ;; since there is nothing in the stack [3,4,2] ;;; dic = {1: 3} [4,2] s =...
LeetCode-Next Greater Element I Description: 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....
Given a circular integer arraynums(i.e., the next element ofnums[nums.length - 1]isnums[0]), returnthenext greater numberfor every element innums. Thenext greater numberof a numberxis the first greater number to its traversing-order next in the array, which means you could search circul...
Leetcode 503. Next Greater Element II 1. Description 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...