【原题】 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.
【分析】 题意为给定一个整形数组，求每个元素的下一个比他大的元素，下一个意为“第一次出现比它大”。 此题类似于leetcode[496]:Next Greater Element |,但是题意改为给定的数组是循环数组而且元素可以出现多次。 采用的stack可以解决此题，思路是用两个stack，一个存放元素，一个存放元素的下标位置。为了解决循环数组的问题，创建一个新数组，将原数组拷贝一份接在后面，即两个原数组相连,长度是原数组额两倍.
【Java】
public class Solution { public int[] nextGreaterElements(int[] nums) { if(nums.length==0 || nums==null) return new int[]{}; int len = nums.length; int[] new_nums = new int[2*len]; for (int i = 0; i < len; i++) { new_nums[i] = nums[i]; new_nums[i+len] = nums[i]; } Stack<Integer> stack = new Stack<Integer>(); int[] ret = new int[2*len]; Arrays.fill(ret, -1); Stack<Integer> idx = new Stack<Integer>(); for (int i = 0;i<new_nums.length;i++) { while(!stack.isEmpty()&&!idx.isEmpty()&& stack.peek()<new_nums[i]){ ret[idx.pop()] = new_nums[i]; stack.pop(); } stack.push(new_nums[i]); idx.push(i); } return Arrays.copyOf(ret, len); } }