# [leetcode]321. 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 k digits. 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]

1：從nums1裡取i個元素組成最大陣列，從nums2裡取k-i個元素組成最大陣列。
2：合併之前結果，得到一個長度為k的最大陣列。
3：對於不同長度分配的情況，比較每次得到的長度為k的最大陣列，返回最大的一個。

``````public class Solution {
public int[] maxNumber(int[] nums1, int[] nums2, int k) {
int n = nums1.length;
int m = nums2.length;
int[] ans = new int[k];
for(int i= Math.max(0, k-m); i<=k && i <= n ; i  ){
int[] candidate = merge(maxArray(nums1,i), maxArray(nums2, k-i), k);
if(greater(candidate, 0, ans, 0)) ans = candidate;
}
return ans;
}

public int[] merge(int[] nums1, int[] nums2, int k){
int[] ans = new int[k];
for(int i=0, j=0, r=0; r<k; r  ){
ans[r] = greater(nums1,i,nums2,j) ? nums1[i  ] : nums2[j  ];
}
return ans;
}

public boolean greater(int[] nums1, int i, int[] nums2, int j){
while(i< nums1.length && j < nums2.length && nums1[i] == nums2[j]){
i  ;
j  ;
}
return j == nums2.length || (i<nums1.length && nums1[i] > nums2[j]);
}

public int[] maxArray(int[] nums, int k){
int n = nums.length;
int[] ans = new int[k];
for(int i=0, j=0; i < n; i  ){
while(n-i> k-j && j > 0 && nums[i] > ans[j-1]) j--;
if(j<k) ans[j  ] = nums[i];
}
return ans;
}
}``````