503. Next Greater Element II

Medium (Trung bình) C++ 🔗 Xem trên LeetCode

📋 Đề Bài

Given a circular integer array nums (i.e., the next element of nums[nums.length - 1] is nums[0]), return the next greater number for every element in nums.

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, return -1 for this number.

 

Example 1:

Input: nums = [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.

Example 2:

Input: nums = [1,2,3,4,3]
Output: [2,3,4,-1,4]

 

Constraints:

  • 1 <= nums.length <= 104
  • -109 <= nums[i] <= 109

🧠 Thuật Toán & Kỹ Thuật

Stack (Ngăn xếp)
⏱️ Thời gian O(n²)
💾 Không gian O(n)

💻 Lời Giải

C++ 0503-next-greater-element-ii.cpp 
class Solution {
public:
    vector<int> nextGreaterElements(vector<int>& nums) {
        const int n = nums.size();
        stack<int> st;
        vector<int> ans(n, -1);
        
        for (int i = 0; i < 2; ++i) {
            for (int j = 0; j < n; ++j) {
                while (!st.empty() and nums[st.top()] < nums[j]) {
                    ans[st.top()] = nums[j];
                    st.pop();
                }
                st.push(j);
            }
        }
        
        return ans;
    }
};