Formatted question description: https://leetcode.ca/all/503.html

503. Next Greater Element II (Medium)

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.

Example 1:

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.

Related Topics:
Stack

Similar Questions:

Solution 1. Monotonic stack

// OJ: https://leetcode.com/problems/next-greater-element-ii/

// Time: O(N)
// Space: O(N)
class Solution {
public:
    vector<int> nextGreaterElements(vector<int>& A) {
        stack<int> s;
        int N = A.size();
        vector<int> ans(N, -1);
        for (int i = 0; i < 2 * N; ++i) {
            int n = A[i % N];
            while (s.size() && A[s.top()] < n) {
                ans[s.top()] = n;
                s.pop();
            }
            s.push(i % N);
        }
        return ans;
    }
};

Java

  • class Solution {
        public int[] nextGreaterElements(int[] nums) {
            int length = nums.length;
            List<Integer> list = new ArrayList<Integer>();
            for (int i = 0; i < length; i++)
                list.add(nums[i]);
            int[] nextGreaterElements = new int[length];
            for (int i = 0; i < length; i++)
                nextGreaterElements[i] = -1;
            Stack<Integer> stack = new Stack<Integer>();
            Stack<Integer> indicesStack = new Stack<Integer>();
            for (int i = 0; i < length; i++) {
                int num = nums[i];
                while (!stack.isEmpty() && stack.peek() < num) {
                    int prevNum = stack.pop();
                    int index = indicesStack.pop();
                    nextGreaterElements[index] = num;
                }
                stack.push(num);
                indicesStack.push(i);
            }
            for (int i = 0; i < length; i++) {
                int num = nums[i];
                while (!stack.isEmpty() && stack.peek() < num) {
                    int prevNum = stack.pop();
                    int index = indicesStack.pop();
                    nextGreaterElements[index] = num;
                }
                stack.push(num);
                indicesStack.push(i);
            }
            return nextGreaterElements;
        }
    }
    
  • // OJ: https://leetcode.com/problems/next-greater-element-ii/
    // Time: O(N)
    // Space: O(N)
    class Solution {
    public:
        vector<int> nextGreaterElements(vector<int>& A) {
            stack<int> s;
            int N = A.size();
            vector<int> ans(N, -1);
            for (int i = 0; i < 2 * N; ++i) {
                int n = A[i % N];
                while (s.size() && A[s.top()] < n) {
                    ans[s.top()] = n;
                    s.pop();
                }
                s.push(i % N);
            }
            return ans;
        }
    };
    
  • class Solution(object):
      def nextGreaterElements(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        ans = [-1] * len(nums)
        n = len(nums)
        stack = []
        nums *= 2
        for i, num in enumerate(nums):
          while stack and nums[stack[-1]] < num:
            top = stack.pop()
            if top < n:
              ans[top] = num
          stack.append(i)
        return ans
    
    

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