[LeetCode] 503. Next Greater Element II 下一个较大的元素 II

2021年09月15日 阅读数:1
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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 circularly to find its next greater number. If it doesn't exist, output -1 for this number.html

Example 1:java

Input: [1,2,1]
Output: [2,-1,2]
Explanation: The first 1's next greater number is 2; 
The number 2 can't find next greater number;
The second 1's next greater number needs to search circularly, which is also 2.

Note: The length of given array won't exceed 10000.python

496. Next Greater Element I 的拓展,这里的数组是循环的,某一个元素的下一个较大值能够在其前面。数组



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]; 
            while (!stack.isEmpty() && nums[stack.peek()] < num)
                next[stack.pop()] = num;
            if (i < n) stack.push(i);
        return next;


def nextGreaterElements(self, nums):
        stack, res = [], [-1] * len(nums)
        for i in range(len(nums)) * 2:
            while stack and (nums[stack[-1]] < nums[i]):
                res[stack.pop()] = nums[i]
        return res 


vector<int> nextGreaterElements(vector<int>& nums) {
        int n = nums.size();
        vector<int> next(n, -1);
        stack<int> s; // index stack
        for (int i = 0; i < n * 2; i++) {
            int num = nums[i % n]; 
            while (!s.empty() && nums[s.top()] < num) {
                next[s.top()] = num;
            if (i < n) s.push(i);
        return next;




[LeetCode] 496. Next Greater Element I 下一个较大的元素 Iblog

[LeetCode] 556. Next Greater Element III 下一个较大的元素 III


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