503. Next Greater Element II 首尾可循环
2021/11/1 6:09:56
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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]
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; }
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