503. Next Greater Element II

题目

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.

我的想法

正向走一遍，再反向走一遍把第一遍没找到的与前面比较
我觉得这道题的test cases真的傻逼透了。找不到返回-1，为什么要有[1,8,-1,-100,-1,222,1111111,-111111]这样的test case？？？？？？？真的写的火大

class Solution {
    public int[] nextGreaterElements(int[] nums) {
        if(nums == null || nums.length <= 0) {
            return new int[0];
        }
        if(nums.length == 1) {
            int[] res = {-1};
            return res;
        }
        int[] res = new int[nums.length];
        Arrays.fill(res, -1);
        int idxRes = 0;
        while(idxRes < nums.length - 1) {
            int idxNums = idxRes + 1;
            while(idxNums < nums.length) {
                if(nums[idxNums] > nums[idxRes]) {
                    res[idxRes] = nums[idxNums];
                    break;
                }
                idxNums++;
            }
            idxRes++;
        }
        
        idxRes = 1;
        while(idxRes < nums.length) {
            int idxNums = 0;
            while(res[idxRes] != -1) {
                idxRes++;
                if(idxRes >= nums.length) {
                    break;
                }
            }
            while(idxNums <= idxRes) {
                if(nums[idxNums] > nums[idxRes]) {
                    res[idxRes] = nums[idxNums];
                    break;
                } else {
                    idxNums++;
                }
            }
            idxRes++;
        }
        return res;
    }
}

解答

Leetcode solution: Stack
很聪明的方法，用一个stack来模拟循环的过程。感觉循环类的题都可以用stack来模拟

class Solution {
    public int[] nextGreaterElements(int[] nums) {
        Stack<Integer> stack = new Stack<>();
        int[] res = new int[nums.length];
        for(int i = nums.length - 1; i >= 0; i--) {
            stack.push(nums[i]);
        }
        for(int i = nums.length - 1; i >= 0; i--) {
            while(!stack.isEmpty() && stack.peek() <= nums[i]) {
                stack.pop();
            }
            if(stack.isEmpty()) {
                res[i] = -1;
            } else {
                res[i] = stack.peek();
            }
            //把自己push进去，便于前一个元素比较
            stack.push(nums[i]);
        }
        return res;
    }
}