Next Greater Element II

Stack

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.

Analysis

此题和Daily Temperatures很像，但是不同之处在于这里允许search circularly，因此常规的做法就是将搜索空间从原先的nums.length，拓展成为2 * nums.length。

其余的要点就是保持一个单调栈，monotonous stack，并且注意数组下标要对nums.length取模，以适应circular的应用场景。

Solution

Monotonous Stack + Circular Array Search (21 ms, faster than 94.64%)

class Solution {
    public int[] nextGreaterElements(int[] nums) {
        int len = nums.length;
        int[] res = new int[len];
        Deque<Integer> stack = new ArrayDeque<>();

        Arrays.fill(res, -1);

        for (int i = 0; i < 2 * len; i++) {
            while (!stack.isEmpty() && nums[i % len] > nums[stack.peek()]) {
                int idx = stack.pop();
                res[idx] = nums[i % len];
            }
            stack.push(i % len);
        }

        return res;
    }
}

• Time complexity : O(n). Only two traversals of the nums array are done. Further, at most 2n elements are pushed and popped from the stack.

• Space complexity : O(n). A stack of size n is used. res array of size n is used.

A little optimization with i < len and checking if stack.isEmpty() then break

class Solution {
    public int[] nextGreaterElements(int[] nums) {
        int len = nums.length;
        int[] res = new int[len];
        Deque<Integer> stack = new ArrayDeque<>();

        Arrays.fill(res, -1);

        for (int i = 0; i < 2 * len; i++) {
            while (!stack.isEmpty() && nums[i % len] > nums[stack.peek()]) {
                int idx = stack.pop();
                res[idx] = nums[i % len];
            }
            if (i < len) {
                stack.push(i);
            }
            if (stack.isEmpty()) {
                break;
            }
        }

        return res;
    }
}