Next Greater Element II Problem


Description

LeetCode Problem 503.

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:

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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:

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

Constraints:

  • 1 <= nums.length <= 10^4
  • -10^9 <= nums[i] <= 10^9


Sample C++ Code

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class Solution {
public:
    vector<int> nextGreaterElements(vector<int>& nums) {
        stack<int> st;
        int idx, len = nums.size();
        vector<int> ans(len, 0);
        
        for (int i = 0; i < 2*len; i ++) {
            idx = i % len;
            while (!st.empty()) {
                if (nums[st.top()] < nums[idx]) {
                    ans[st.top()] = nums[idx];
                    st.pop();
                } else {
                    break;
                }
            }
            if (i < len)
                st.push(idx);
        }
        while (!st.empty()) {
            ans[st.top()] = -1;
            st.pop();
        }
        return ans;
    }
};




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