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...
publicintnextGreaterElement(intn) { char[] number = (n +"").toCharArray(); inti, j; // I) Start from the right most digit and // find the first digit that is // smaller than the digit next to it. for(i = number.length-1; i >0; i--) if(number[i-1] < number[i]) br...
题目链接: https://leetcode-cn.com/problems/next-greater-element-iii难度:中等 通过率:26.3% 题目描述:给定一个 32位 正整数 n ,你需要找到最小的 32位 整数,其与 n 中存在的位数完全相同,并且其值大于n…
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...
leetcode556. Next Greater Element III 题目要求 Given a positive32-bitintegern, you need to find the smallest32-bitinteger which has exactly the same digits existing in the integernand is greater in value than n. If no such positive32-bitinteger exists, you need to return -1....
LeetCode556 下一个更大元素https://leetcode-cn.com/problems/next-greater-element-iii/ 二、滑动窗口 LeetCode3 Medium 无重复字符的最长子串https://leetcode-cn.com/problems/longest-substring-without-repeating-characters/ LeetCode76 Hard最小覆盖子串https://leetcode-cn.com/problems/minimum-window-subst...
[leetcode] 503. Next Greater Element II Description 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...
Given a circular integer arraynums(i.e., the next element ofnums[nums.length - 1]isnums[0]), returnthenext greater numberfor every element innums. Thenext greater numberof a numberxis the first greater number to its traversing-order next in the array, which means you could search circul...
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 ...
借助stack,遍历nums,如果stack为空或者nums[stack[-1]] >= nums[i],将i压入stack,表示nums[i]还未找到它的下一个greater element。当nums[stack[-1]]<nums[i]时,表示nums[i]即为nums[stack[-1]]的下一个较大元素,因此从stack中取出该索引index,将nums[i]存入result[index]中。由于nums是循环的,因此...