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.
// Brute Force
vector<int> nextGreaterElements(vector<int>& nums) { // time: O(n^2); space: O(n)
int n = nums.size();
vector<int> res(n, -1);
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (nums[(i + j) % n] > nums[i]) {
res[i] = nums[(i + j) % n];
break;
}
}
}
return res;
}
// Stack
vector<int> nextGreaterElements(vector<int>& nums) { // time: O(n); space: O(n)
int n = nums.size();
vector<int> res(n, -1);
stack<int> st;
for (int i = 0; i < 2 * n; ++i) {
int num = nums[i % n];
while (!st.empty() && nums[st.top()] < num) {
res[st.top()] = num;
st.pop();
}
if (i < n) st.push(i);
}
return res;
}