问题：

给定长度分别为 m 和 n 的两个数组，其元素由 0-9 构成，表示两个自然数各位上的数字。现在从这两个数组中选出 k (k <= m + n) 个数字拼接成一个新的数，要求从同一个数组中取出的数字保持其在原数组中的相对顺序。

求满足该条件的最大数。结果返回一个表示该最大数的长度为 k 的数组。

说明: 请尽可能地优化你算法的时间和空间复杂度。

示例 1:

输入:
nums1 = 
[3, 4, 6, 5]

nums2 = 
[9, 1, 2, 5, 8, 3]

k = 
5

输出:
[9, 8, 6, 5, 3]
示例 2:

输入:
nums1 = 
[6, 7]

nums2 = 
[6, 0, 4]

k = 
5

输出:
[6, 7, 6, 0, 4]
示例 3:

输入:
nums1 = 
[3, 9]

nums2 = 
[8, 9]

k = 
3

输出:
[9, 8, 9]

解答思路：

以下是使用 Java 实现拼接最大数的示例代码：

import java.util.Arrays;import java.util.Comparator;import java.util.PriorityQueue;class Solution {public int[] maxNumber(int[] nums1, int[] nums2, int k) {int m = nums1.length;int n = nums2.length;int[] result = new int[k];Arrays.fill(result, -1);// 从数组 nums1 中选择 i 个数字，从数组 nums2 中选择 k - i 个数字for (int i = Math.max(0, k - n); i <= Math.min(k, m); i++) {int[] subsequence1 = getMaxSubsequence(nums1, i);int[] subsequence2 = getMaxSubsequence(nums2, k - i);int[] merged = merge(subsequence1, subsequence2);if (isGreater(merged, result, 0, 0)) {System.arraycopy(merged, 0, result, 0, k);}}return result;}// 获取数组的最大子序列，长度为 lenprivate int[] getMaxSubsequence(int[] nums, int len) {int[] stack = new int[len];int top = -1;for (int i = 0; i < nums.length; i++) {while (top >= 0 && stack[top] < nums[i] && top + nums.length - i > len) {top--;}if (top < len - 1) {stack[++top] = nums[i];}}return stack;}// 合并两个子序列private int[] merge(int[] subsequence1, int[] subsequence2) {int[] merged = new int[subsequence1.length + subsequence2.length];int i = 0, j = 0, k = 0;while (i < subsequence1.length && j < subsequence2.length) {if (subsequence1[i] > subsequence2[j]) {merged[k++] = subsequence1[i++];} else if (subsequence1[i] < subsequence2[j]) {merged[k++] = subsequence2[j++];} else {int p = i + 1, q = j + 1;while (p < subsequence1.length && q < subsequence2.length && subsequence1[p] == subsequence2[q]) {p++;q++;}if (q == subsequence2.length || (p < subsequence1.length && subsequence1[p] > subsequence2[q])) {merged[k++] = subsequence1[i++];} else {merged[k++] = subsequence2[j++];}}}while (i < subsequence1.length) {merged[k++] = subsequence1[i++];}while (j < subsequence2.length) {merged[k++] = subsequence2[j++];}return merged;}// 比较两个整数数组，如果 num1 更大，则返回 true，否则返回 falseprivate boolean isGreater(int[] num1, int[] num2, int index1, int index2) {while (index1 < num1.length && index2 < num2.length) {if (num1[index1]!= num2[index2]) {return num1[index1] > num2[index2];}index1++;index2++;}return index1!= num1.length;}public static void main(String[] args) {Solution solution = new Solution();int[] nums1 = {3, 4, 6, 5};int[] nums2 = {9, 1, 2, 5, 8, 3};int k = 5;int[] result = solution.maxNumber(nums1, nums2, k);System.out.println(Arrays.toString(result));}}

上述示例代码中，定义了一个名为'Solution'的类。'maxNumber'方法用于拼接两个数组的最大数，通过遍历从两个数组中选择不同长度的数字，得到最大子序列，并利用辅助方法'merge'、'getMaxSubsequence'和'isGreater'进行合并、获取子序列和比较。'merge'方法用于合并两个数组，利用双指针进行比较和合并。'getMaxSubsequence'方法用于获取最大子序列，通过单调栈保存当前最大数字。'isGreater'方法用于比较两个整数数组的大小。

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