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...
Given a circular integer arraynums(i.e., the next element ofnums[nums.length - 1]isnums[0]), returnthe next greater number for every element innums. The next greater number of a numberxis the first greater number to its traversing-order next in the array, which means you could search...
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...
复制 classSolution{public:vector<int>nextGreaterElements(vector<int>&nums){int n=nums.size();stack<int>st;vector<int>res(n,-1);for(int i=2*n-2;i>=0;--i){while(!st.empty()&&nums[i%n]>=st.top()){st.pop();}res[i%n]=st.empty()?-1:st.top();st.push(nums[i%n]);}r...
[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...
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...
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, 需要index来填res. Time Complexity: O(n), n = nums.length. ...
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...
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...
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