[LeetCode] Create Maximum Number

Given two arrays of length m and n with digits 0-9 representing two numbers. Create the maximum number of length k <= m + n from digits of the two. The relative order of the digits from the same array must be preserved. Return an array of the kdigits. You should try to optimize your time and space complexity.

Example 1:

nums1 = [3, 4, 6, 5]
nums2 = [9, 1, 2, 5, 8, 3]
k = 5
return [9, 8, 6, 5, 3]

Example 2:

nums1 = [6, 7]
nums2 = [6, 0, 4]
k = 5
return [6, 7, 6, 0, 4]

Example 3:

nums1 = [3, 9]
nums2 = [8, 9]
k = 3
return [9, 8, 9]

这题挺恶心的，我的做法不太好，勉强通过case（第二遍刷的时候用的Stack容器，而不是数组，变慢了好多）

public class Solution {

    public int[] maxNumber(int[] nums1, int[] nums2, int k) {
    	int[] res=null;
    	for(int i=Math.max(k-nums2.length,0);i<=Math.min(k,nums1.length);i++){
    		int[] a=maxNumber(nums1, i);
    		int[] b=maxNumber(nums2, k-i);
    		int[] re=merge(a,b);
    		if(res==null) res=re;
    		else res=compareTo(res,0,re,0)>0?res:re;
    	}
    	return res;
    }
    public int[] merge(int[] nums1,int[] nums2){
    	int[] re=new int[nums1.length+nums2.length];
    	for(int index=0,i=0,j=0;index<nums1.length+nums2.length;index++){
    		if(i>=nums1.length){
    			re[index]=nums2[j++];
    		}else if(j>=nums2.length){
    			re[index]=nums1[i++];
    		}else{
    			if(compareTo(nums1, i, nums2, j)>0) re[index]=nums1[i++];
    			else re[index]=nums2[j++];
    		}
    	}
    	return re;
    }
    public int[] maxNumber(int[] nums, int k) {
		 if(nums.length<k) return null;
		 int[] stack=new int[1000];
		 int index=0;
		 for(int i=0;i<nums.length;i++){
			 while(index!=0&&nums.length-i+index>k&&stack[index-1]<nums[i]){
				index--;
			 }				 
			 stack[index++]=nums[i];
		 }
		 return Arrays.copyOfRange(stack,0,k);
	 }
    public int compareTo(int[] nums1,int i,int[] nums2,int j){
    	while(i<nums1.length&&j<nums2.length){
    		if(nums1[i]==nums2[j]){
    			i++;j++;
    		}else{
    			return nums1[i]-nums2[j];
    		}
    	}
    	if(nums1.length==i&&nums2.length==j) return 0;
    	else if(i<nums1.length) return 1;
    	else return -1;
    }
    
 
}