Can you solve this real interview question? Next Greater Element II - Given a circular integer array nums (i.e., the next element of nums[nums.length - 1] is nums[0]), return the next greater number for every element in nums. The next greater number of
[LeetCode] 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 arr...
503. Next Greater Element II Problem: 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 ...
LeetCode:Next Greater Element II 503. 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 traver...
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...
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[-1]]<nums[i]: res[stack.pop()] = nums[i]...
详见:https://leetcode.com/problems/next-greater-element-ii/description/ C++: classSolution{public:vector<int>nextGreaterElements(vector<int>&nums){intn=nums.size();vector<int>res(n,-1);for(inti=0;i<n;++i){for(intj=i+1;j<i+n;++j){if(nums[j%n]>nums[i]){res[i]=nums[j%n]...
publicintnextGreaterElement(int n){String value=String.valueOf(n);char[]digits=value.toCharArray();int i=digits.length-1;//找到小于右侧任意值的第一个正整数while(i>0){if(digits[i-1]<digits[i]){break;}i--;}if(i==0){return-1;}//找到该整数右侧大于该整数的最小整数int maxIndex=i,...
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 nums2. The Next Greater Number of a number x in nums1...
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): ...