https://leetcode.com/problems/create-maximum-number/

思路有点像merge sort.参考https://leetcode.com/discuss/75772/share-my-python-solution-with-explanation

To create the max number from num1 and nums2 with k elements, we assume the final result combined by i numbers (denotes as left) from num1 and j numbers (denotes as right) from nums2, where i+j==k.

Obviously, left and right must be the maximum possible number in num1 and num2 respectively. i.e. num1 = [6,5,7,1] and i == 2, then left must be [7,1]. 这里为什么不是[7,6]呢，因为要preserve the relative order

The final result is the maximum possible merge of all left and right.

So there're 3 steps:

- iterate i from 0 to k.
- find max number from num1, num2 by select i , k-i numbers, denotes as left, right
- find max merge of left, right

function maxSingleNumber select i elements from num1 that is maximum. The idea find the max number one by one. i.e. assume nums [6,5,7,1,4,2], selects = 3. 1st digit: find max digit in [6,5,7,1], the last two digits [4, 2] can not be selected at this moment.因为至少要留2个数，否则接下来就没有数可以选了。 2nd digits: find max digit in [1,4], since we have already selects 7, we should consider elements after it, also, we should leave one element out. 3rd digits: only one left [2], we select it. and function output [7,4,2]

function mergeMax find the maximum combination of left, and right.

class Solution(object):
    def maxNumber(self, nums1, nums2, k):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :type k: int
        :rtype: List[int]
        """
        n, m= len(nums1),len(nums2)
        ret = [0] * k
        for i in range(0, k+1):
            j = k - i
            if i > n or j > m: continue
            left = self.maxSingleNumber(nums1, i)
            right = self.maxSingleNumber(nums2, j)
            num = self.mergeMax(left, right)
            ret = max(num, ret)
        return ret


    def mergeMax(self, nums1, nums2):
        ans = []
        while nums1 or nums2:
            if nums1 > nums2:
                ans += nums1[0],
                nums1 = nums1[1:]
            else:
                ans += nums2[0],
                nums2 = nums2[1:]
        return ans

    def maxSingleNumber(self, nums, selects):
        n = len(nums)
        ret = [-1]
        if selects > n : return ret
        tmp1 = [0]
        while selects > 0:
            start = ret[-1] + 1 #search start
            #search end,因为我们要选selects个值，所以搜索最大值的范围，最多也就到n-selects + 1
            end = n-selects + 1
            #这里要注意，ret保存的只是index，所以这里append的是max(nums[start:end])的index，所以不能直接用max(nums[start:end])
            ret.append( max(range(start, end), key = nums.__getitem))
            selects -= 1
        ret = [nums[item] for item in ret[1:]]
        return ret