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]

一刷
题解：
一种很简单的思路，分别从左边取出[0-k]个元素，从右边取出[k-0]个元素，组成长度为k的新array，并比较选择其中一个最大的。

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;
    }

    private 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]);
    }


//从array中取出k个最大的元素，并保留原来的顺序
    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 + j > k && j > 0 && ans[j - 1] < nums[i]) j--;
            if (j < k) ans[j++] = nums[i];
        }
        return ans;
    }
}