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:
1
Input: [1,2,1]
2
3
Output: [2,-1,2]
4
5
Explanation:
6
7
The first 1's next greater number is 2;
8
The number 2 can't find next greater number;
9
The second 1's next greater number needs to search circularly, which is also 2.
Copied!
Note:The length of given array won't exceed 10000.

2. Implementation

思路:这道题和Next Greater Element I的区别就在于这里数组是一个环,所以数组最后的数组的Greater Next Element可能在它前面。方法就是拆环成线,另外和一个不同点是Next Greater Element I的输入是两个数组,而这里只有一个数组,所以stack存的是数组的Index而不是数组的元素
1
class Solution {
2
public int[] nextGreaterElements(int[] nums) {
3
int n = nums.length;
4
int[] res = new int[n];
5
Arrays.fill(res, -1);
6
7
Stack<Integer> stack = new Stack<>();
8
9
for (int i = 0; i < 2 * n; i++) {
10
int num = nums[i % n];
11
while (!stack.isEmpty() && nums[stack.peek()] < num) {
12
res[stack.pop()] = num;
13
}
14
15
if (i < n) {
16
stack.push(i);
17
}
18
}
19
20
return res;
21
}
22
}
Copied!

3. Time & Space Complexity

时间和空间复杂度都是O(n)