求解子问题，合并子问题，求解子问题的方法为单调栈

代码：

1. maxSubsequence 函数来求单调栈
2. lexicographicalLess 函数来比较两个数组的大小
3. merge 函数来合并两个子数组
4. maxNumber 函数，来分配子问题，并维护答案最大值
5. 分配子问题，即，从数组一中取多少个数字，从数组二中取多少个数字，for 循环求解子问题即可

func maxSubsequence(a []int, k int) (s []int) {
    for i, v := range a {
        for len(s) > 0 && len(s)+len(a)-1-i >= k && v > s[len(s)-1] {
            s = s[:len(s)-1]
        }
        if len(s) < k {
            s = append(s, v)
        }
    }
    return
}

func lexicographicalLess(a, b []int) bool {
    for i := 0; i < len(a) && i < len(b); i++ {
        if a[i] != b[i] {
            return a[i] < b[i]
        }
    }
    return len(a) < len(b)
}

func merge(a, b []int) []int {
    merged := make([]int, len(a)+len(b))
    for i := range merged {
        if lexicographicalLess(a, b) {
            merged[i], b = b[0], b[1:]
        } else {
            merged[i], a = a[0], a[1:]
        }
    }
    return merged
}

func maxNumber(nums1, nums2 []int, k int) (res []int) {
    start := 0
    if k > len(nums2) {
        start = k - len(nums2)
    }
    for i := start; i <= k && i <= len(nums1); i++ {
        s1 := maxSubsequence(nums1, i)
        s2 := maxSubsequence(nums2, k-i)
        merged := merge(s1, s2)
        if lexicographicalLess(res, merged) {
            res = merged
        }
    }
    return
}