503. Next Greater Element II-优快云博客

该博客介绍了如何在循环数组中寻找每个元素的下一个更大元素，提供两种不同的算法方法。方法一是利用栈和队列，分别存储降序和升序子集，方法二是通过一次循环和栈来保存降序序列的索引。文章详细解释了每种方法的思路，并给出了相应的代码实现。

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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.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/next-greater-element-ii
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

在一个循环列表中，找到每个数字中，下一个比他大的数字

方法一：

用一个栈，一个队列。栈放置一个降序的子集，队列放一个升序的子集。


class Solution {
    public int[] nextGreaterElements(int[] nums) {
        int []ans = new int[nums.length];
// 保存降序子集
        Stack<Integer> num = new Stack<>();
        Stack<Integer> index = new Stack<>();
// 保存升序子集
        Queue<Integer> upNum = new ArrayDeque<>();
        int maxValue = Integer.MIN_VALUE;
        for (int i = 0; i < nums.length; i++) {
            if (num.empty()) {
                num.push(nums[i]);
                index.push(i);
                upNum.add(nums[i]);
                maxValue = nums[i];
            } else {
                while (!num.empty() && nums[i] > num.peek()) {
                    ans[index.peek()] = nums[i];
                    num.pop();
                    index.pop();
                }
                num.push(nums[i]);
                index.push(i);
                if (nums[i] > maxValue) {
                    upNum.add(nums[i]);
                    maxValue = nums[i];
                }
            }
        }
// 降序中，如果还有没有标记的，就再从升序队列中找到。
        while (!num.empty()) {
            int curNume = num.pop();
            int curIndex = index.pop();
            if (curNume == maxValue) {
                ans[curIndex] = -1;
                continue;
            }
            while (!upNum.isEmpty()) {
                if (curNume < upNum.peek()) {
                    ans[curIndex] = upNum.peek();
                    break;
                } else {
                    upNum.poll();
                }
            }
        }
        return ans;
    }
}

方法二：

循环是2倍的数组长度，用栈来保存降序序列的index。

class Solution {
public int[] nextGreaterElements(int[] nums) {
        int n = nums.length, next[] = new int[n];
        Arrays.fill(next, -1);
        Stack<Integer> stack = new Stack<>(); // index stack
        for (int i = 0; i < n * 2; i++) {
            int num = nums[i % n];
            while (!stack.isEmpty() && nums[stack.peek()] < num)
                next[stack.pop()] = num;
// 遍历第一轮，把没有找到比他大的数字放起来，所以stack里面的数据都是降序的。然后再进行第二轮遍历
            if (i < n) stack.push(i);
        }  
        return next;
    }
}