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...
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
Can you solve this real interview question? Next Greater Element I - The next greater element of some element x in an array is the first greater element that is to the right of x in the same array. You are given two distinct 0-indexed integer arrays num
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 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。 ......
LeetCode——496. 下一个更大元素 I[Next Greater Element I][简单]——分析及代码[C++] 一、题目 二、分析及代码 1. 单调栈 + 哈希表 (1)思路 (2)代码 (3)结果 三、其他 一、题目 给你两个 没有重复元素 的数组 nums1 和 nums2 ,其中nums1&nbs...Leet...
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): ...
Leetcode 556. Next Greater Element III 2. Solution **解析:**Version 1,先将数字n变为字符数组,要找最小的大于n的数,则应该从右往左开始,依次寻找第i位字符右边的大于当前字符的最小数字,然后互换二者位置,由于新数字的第i位字符大于n中的第i位字符,因此新数字i位之后的字符应该从小到大排列,这样可以...