Problem
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
Solution
/**
* @param {number[]} nums
* @return {number[]}
*/
var nextGreaterElements = function(nums) {
var map = Array(nums.length);
var stack = [];
for (var i = nums.length * 2 - 1; i >= 0; i--) {
var index = i % nums.length;
while (stack.length && stack[stack.length - 1] <= nums[index]) stack.pop();
map[index] = stack.length ? stack[stack.length - 1] : -1;
stack.push(nums[index]);
}
return map;
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).