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...
Java publicint[] nextGreaterElement(int[] nums1,int[] nums2) { Stack<Integer> stack =newStack<>(); HashMap<Integer, Integer> map =newHashMap<>();for(inti=nums2.length -1; i >=0; i--) {while(!stack.isEmpty() && stack.peek() < nums2[i]) { stack.pop(); }if(!stack.is...
publicint[]nextGreaterElement(int[]findNums,int[]nums){int[]ans=newint[findNums.length];Map<Integer,Integer>map=newHashMap<>();Stack<Integer>stack=newStack<>();for(int num:nums){while(!stack.isEmpty()&&stack.peek()<num)map.put(stack.pop(),num);stack.push(num);}for(int i=0;i<...
this stack we are gonna to use is actually a mono decrease stack, and we store the element of nums1 as the key, and the final results as the value corresponding to that element in nums1. public int[] nextGreaterElement(int[] findNums, int[] nums) { Map<Integer, Integer> map = ne...
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。
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[...
题目描述: LeetCode 496. Next Greater Element I 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 nums