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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;
    }
};

  • 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;
    }
}

############

class Solution {
    public int[] nextGreaterElements(int[] nums) {
        int n = nums.length;
        int[] ans = new int[n];
        Arrays.fill(ans, -1);
        Deque<Integer> stk = new ArrayDeque<>();
        for (int i = 0; i < (n << 1); ++i) {
            while (!stk.isEmpty() && nums[stk.peek()] < nums[i % n]) {
                ans[stk.pop()] = nums[i % n];
            }
            stk.push(i % n);
        }
        return ans;
    }
}

  • // 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:
    def nextGreaterElements(self, nums: List[int]) -> List[int]:
        n = len(nums)
        ans = [-1] * n
        stk = []
        for i in range(n << 1):
            while stk and nums[stk[-1]] < nums[i % n]:
                ans[stk.pop()] = nums[i % n]
            stk.append(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


  • func nextGreaterElements(nums []int) []int {
	n := len(nums)
	ans := make([]int, n)
	for i := range ans {
		ans[i] = -1
	}
	var stk []int
	for i := 0; i < (n << 1); i++ {
		for len(stk) > 0 && nums[stk[len(stk)-1]] < nums[i%n] {
			ans[stk[len(stk)-1]] = nums[i%n]
			stk = stk[:len(stk)-1]
		}
		stk = append(stk, i%n)
	}
	return ans
}

  • function nextGreaterElements(nums: number[]): number[] {
    const stack: number[] = [],
        len = nums.length;
    const res: number[] = new Array(len).fill(-1);
    for (let i = 0; i < 2 * len - 1; i++) {
        const j = i % len;
        while (stack.length !== 0 && nums[stack[stack.length - 1]] < nums[j]) {
            res[stack[stack.length - 1]] = nums[j];
            stack.pop();
        }
        stack.push(j);
    }
    return res;
}


  • /**
 * @param {number[]} nums
 * @return {number[]}
 */
var nextGreaterElements = function (nums) {
    const n = nums.length;
    let stk = [];
    let ans = new Array(n).fill(-1);
    for (let i = 0; i < n << 1; i++) {
        const j = i % n;
        while (stk.length && nums[stk[stk.length - 1]] < nums[j]) {
            ans[stk.pop()] = nums[j];
        }
        stk.push(j);
    }
    return ans;
};

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