503 Next Greater Element II

# 1. Question

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:
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Input: [1,2,1]
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Output: [2,-1,2]
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Explanation:
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The first 1's next greater number is 2;
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The number 2 can't find next greater number;
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The second 1's next greater number needs to search circularly, which is also 2.
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Note:The length of given array won't exceed 10000.

# 2. Implementation

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class Solution {
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public int[] nextGreaterElements(int[] nums) {
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int n = nums.length;
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int[] res = new int[n];
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Arrays.fill(res, -1);
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Stack<Integer> stack = new Stack<>();
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for (int i = 0; i < 2 * n; i++) {
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int num = nums[i % n];
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while (!stack.isEmpty() && nums[stack.peek()] < num) {
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res[stack.pop()] = num;
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}
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if (i < n) {
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stack.push(i);
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}
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}
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return res;
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}
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}
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