数字2找不到下一个更大的数; 第二个1的下一个最大的数需要循环搜索,结果也是2。 来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/next-greater-element-ii 2. 解题 2.1 朴素解法 模拟题意查找 代码语言:javascript 复制 classSolution{public:vector<int>nextGreaterElements(vector<int>&nums){...
classSolution {public: vector<int> nextGreaterElements(vector<int>&nums) {intn =nums.size(); vector<int> res(n, -1);//技巧:提前都初始化为-1,减少后面重新赋值的步骤for(inti =0; i <2* n; ++i) {intnum = nums[i %n];while(!st.empty() && nums[st.top()] <num) { res[st.t...
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 traversing-order next in the array, w...
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...
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...
Leetcode 503. Next Greater Element II 1. Description 2. Solution **解析:**Version 1,由于元素不是唯一的,需要循环查找,因此先将nums复制一遍,通过循环每次都查找当前元素之后的n-1位数字。Version 2通过使用栈来寻找满足条件的结果,栈中保持是数字的索引位置,由于需要循环查找,因此需要查找两次nums,并且第二...
classSolution{publicint[]nextGreaterElements(int[]nums){if(nums==null||nums.length==0)returnnums;// 存储数字idxStack<Integer>stack=newStack<>();// key: 数字idx,value:next Greater ElementMap<Integer,Integer>map=newHashMap<>();for(inti=0;i<2*nums.length;i++){while(!stack.isEmpty()&&...
leetcode503. next greater element ii 【m】用户正在刷题磕cp 29 播放 · 0 弹幕 【300题刷题挑战】leetcode力扣1 两数之和 twosum 第一百四十四题 | 哈希表 soulbit花云田 69 播放 · 0 弹幕 09 - 面试题:用队列实现栈&用栈实现队列爱喝柠檬茶的喵酱 531 播放 · 1 弹幕 【leet...
2. 3. 4. 5. Note: The length of given array won't exceed 10000. 题解: 是Next Greater Element I进阶版. 一般这类带环的一种常见做法是加长一倍. 这里就是把nums 走两边. 没用HashMap<Integer, Integer>是因为nums里会有duplicate. 这里的Stack<Integer> stk 存的是element index 而不是element, 需...
public int[] nextGreaterElements(int[] nums) { int n = nums.length, next[] = new int[n]; Arrays.fill(next, -1); Stack<Integer> stack = new Stack<>(); // index stack for (int i = 0; i < n * 2; i++) { int num = nums[i % n]; ...