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.
496. Next Greater Element I 的拓展，这里的数组是循环的，某一个元素的下一个较大值可以在其前面。
解法：栈，与496的不同是：循环是2倍的数组长度，用栈来保存降序序列的index。
Java:
public int[] nextGreaterElements(int[] nums) {
int n = nums.length, next[] = new int[n];
Arrays.fill(next, -1);
Stack<Integer> stack = new Stack<>(); // index stack
for (int i = 0; i < n * 2; i++) {
int num = nums[i % n];
while (!stack.isEmpty() && nums[stack.peek()] < num)
next[stack.pop()] = num;
if (i < n) stack.push(i);
}
return next;
}
Python:
def nextGreaterElements(self, nums):
stack, res = [], [-1] * len(nums)
for i in range(len(nums)) * 2:
while stack and (nums[stack[-1]] < nums[i]):
res[stack.pop()] = nums[i]
stack.append(i)
return res
C++:
vector<int> nextGreaterElements(vector<int>& nums) {
int n = nums.size();
vector<int> next(n, -1);
stack<int> s; // index stack
for (int i = 0; i < n * 2; i++) {
int num = nums[i % n];
while (!s.empty() && nums[s.top()] < num) {
next[s.top()] = num;
s.pop();
}
if (i < n) s.push(i);
}
return next;
}
类似题目：
[LeetCode] 496. Next Greater Element I 下一个较大的元素 I
[LeetCode] 556. Next Greater Element III 下一个较大的元素 III