503. Next Greater Element II
Description Submission Solutions
- Total Accepted: 3567
- Total Submissions: 7703
- Difficulty: Medium
- Contributors: love_FDU_llp
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.
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【题目分析】
这个题目与前一个题目类似，不同的是数组中的元素可以组成一个圈，同样是找到某个元素右边第一个比他大的元素。
【思路】
这个题目同样是通过栈来解决，不过相比前一个题目，这个题目中的栈存储的是数组元素的下标。并且数组元素组成一个圈的话为了找出所有元素右边的最大值，我们需要把数组遍历两遍就足够了。由于在第一遍中所有元素都已经入栈，因此在第二遍的时候不需要再将元素入栈。
【java代码】
1 public class Solution { 2 public int[] nextGreaterElements(int[] nums) { 3 int len = nums.length; 4 int[] res = new int[len]; 5 Arrays.fill(res, -1); 6 Stack<Integer> stack = new Stack<>(); 7 8 for(int i = 0; i < 2*len; i++) { 9 int num = nums[i%len]; 10 while(!stack.isEmpty() && nums[stack.peek()] < num) { 11 res[stack.pop()] = num; 12 } 13 if(i < len) stack.push(i); 14 } 15 16 return res; 17 } 18 }