Next Greater Element II – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
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
C++ Next Greater Element II LeetCode Solution
vector<int> nextGreaterElements(vector<int>& A) {
int n = A.size();
vector<int> stack, res(n, -1);
for (int i = 0; i < n * 2; ++i) {
while (stack.size() && A[stack.back()] < A[i % n]) {
res[stack.back()] = A[i % n];
stack.pop_back();
}
stack.push_back(i % n);
}
return res;
}
Java Next Greater Element II LeetCode Solution
public int[] nextGreaterElements(int[] A) {
int n = A.length, res[] = new int[n];
Arrays.fill(res, -1);
Stack<Integer> stack = new Stack<>();
for (int i = 0; i < n * 2; i++) {
while (!stack.isEmpty() && A[stack.peek()] < A[i % n])
res[stack.pop()] = A[i % n];
stack.push(i % n);
}
return res;
}
Python 3 Next Greater Element II LeetCode Solution
def nextGreaterElements(self, A):
stack, res = [], [-1] * len(A)
for i in range(len(A)) * 2:
while stack and (A[stack[-1]] < A[i]):
res[stack.pop()] = A[i]
stack.append(i)
return res
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