Problem 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.
Solution
This question is based on: Next Greater Element I The trick is to add all elements in the stack instead of just the last elements (from right to left) because the question mentions that we can treat is as a cirular array.
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class Solution {
public int[] nextGreaterElements(int[] nums) {
int n = nums.length;
int ans[] = new int[n];
Stack<Integer> stack = new Stack<>();
for(int i=n-1; i>=0; i--){
stack.push(nums[i]);
}
for(int i=n-1; i>=0; i--){
int current = nums[i];
while(!stack.isEmpty() && current >= stack.peek()){
stack.pop();
}
if(stack.isEmpty()){
ans[i] = -1;
stack.push(current);
}else{
ans[i] = stack.peek();
stack.push(current);
}
}
return ans;
}
}