【1】,【1】背后看到的第一个元素是2,因此next greater为2; 【2】,【2】背后看到的第一个元素是4,因此next greater为4;(蓝色2) 在代码中为了记录某个值背后没被挡住的第一个元素,这里使用栈stack来记录,入栈出栈规则:当前值比栈中元素大时(证明被挡住了),将该值出栈,直至栈空或当前值小于栈顶元素,此时...
Explanation: For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1. For number 1 in the first array, the next greater number for it in the second array is 3. For number 2 in the first array, there is no next greater n...
For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1. For number 1 in the first array, the next greater number for it in the second array is 3. For number 2 in the first array, there is no next greater number for it...
For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1. For number 1 in the first array, the next greater number for it in the second array is 3. For number 2 in the first array, there is no next greater number for it...
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 is the first greater number to its ...
Explanation:For number 2 in the first array, the next greater number for it in the second array is 3. For number 4 in the first array, there is no next greater number for it in the second array, so output -1. question 解法:
implSolution{pubfnnext_greater_element(nums1:Vec<i32>,nums2:Vec<i32>)->Vec<i32>{letmutans=vec![-1;nums1.len()];letmutst=vec![];letmutcount=0;letmutmap=std::collections::HashMap::new();for(i,&x)innums1.iter().enumerate(){map.insert(x,i);};// 需要使用哈希表来解决两个数组...
Can you solve this real interview question? Next Greater Numerically Balanced Number - An integer x is numerically balanced if for every digit d in the number x, there are exactly d occurrences of that digit in x. Given an integer n, return the smallest
For number 2 in the first array, there is no next greater number for it in the second array, so output -1. My solution: class Solution { public int[] nextGreaterElement(int[] nums1, int[] nums2) { int[] res = new int[nums1.length]; ...
leetcode503. Next Greater Element II 题目要求 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 ...