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;}
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,这样可以保...
[LeetCode] 496. Next Greater Element I 下一个较大的元素 I [LeetCode] 503. Next Greater Element II 下一个较大的元素 II All LeetCode Questions List 题目汇总
解法一: classSolution {public:intnextGreaterElement(intn) {stringstr =to_string(n);intlen = str.size(), i = len -1;for(; i >0; --i) {if(str[i] > str[i -1])break; }if(i ==0)return-1;for(intj = len -1; j >= i; --j) {if(str[j] > str[i -1]) { swap(st...
题目描述: LeetCode 556. Next Greater Element III Given a positive 32-bit integer n, you need to find the smallest 32-bit integer which has exactly the same digits existing in the integer n and is greater in value than n. If no such positive 32-bit integ
public int nextGreaterElement(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]) { ...
LeetCode 556. Next Greater Element III https://leetcode.com/problems/next-greater-element-iii/description/ 题目: Given a positive 32-bit integer n, you need to find the smallest 32-bit integer which has exactly the same digits existing in the integer nand is greater in value than n. If...
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——503. 下一个更大元素 II[Next Greater Element II][中等]——分析及代码[Java] 一、题目 二、分析及代码 1. 单调栈 (1)思路 (2)代码 (3)结果 三、其他 一、题目 给定一个循环数组(最后一个元素的下一个元素是数组的第一个元素),输出每个元素的下一个更大元素。数字 x 的下一个更大...
public int[] nextGreaterElement(int[] findNums, int[] nums) { Map<Integer, Integer> map = new HashMap<>(); // map from x to next greater element of x Stack<Integer> stack = new Stack<>(); for (int num : nums) { //pay attention, we use nums2 to constuact our map, and ...