Description
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]
Constraints:
1 <= nums.length <= 104-109 <= nums[i] <= 109
Code
Monotonic Stack
Time Complexity: , Space Complexity:
Trick：繞兩圈，real_i = i % n
剩下的就都和 Next Greater Element I 一樣了。
class Solution {
public:
vector<int> nextGreaterElements(vector<int>& nums) {
unordered_map<int, int> mp;
int n = nums.size();
vector<int> res(n, -1);
stack<int> st;
for(int i = 0; i < 2 * n; i++) {
int real_i = i % n;
while(!st.empty() && nums[st.top()] < nums[real_i]) {
res[st.top()] = nums[real_i];
st.pop();
}
st.push(real_i);
}
return res;
}
};