拼接最大数（单调栈）

Leetcode 每日一题
题目链接： 321. 拼接最大数
难度： 困难
解题思路： 首先确定每个数组取的字符的数量。满足两者x + y = k。其中x和y分别是两个字符数组的中取值的数量。分别从这两个大的字符数组中取其中的最大的子序列。通过对这些子序列的字典序比较，获取到最大的拼接数。在获取拼接数的时候可以使用单调栈进行操作。保证栈顶存的一直是字典序相对较大的序列。
题解：

class Solution:
    def maxNumber(self, nums1: List[int], nums2: List[int], k: int) -> List[int]:

        # 比较两个剩下的串的字典序和长度大小
        def compare(list1, idx1, list2, idx2):
            len1, len2 = len(list1), len(list2)

            while idx1 < len1 and idx2 < len2:
                d = (list1[idx1] - list2[idx2])
                if d != 0:
                    return d
                idx1 += 1
                idx2 += 1
            
            return (len1 - idx1) - (len2 - idx2)
        # 单调栈
        def monoStack(list1, list2) -> List[int]:
            len1, len2 = len(list1), len(list2)
            stack = list()
            i, j = 0, 0
            while i < len1 and j < len2:
                if compare(list1, i, list2, j) > 0:
                    stack.append(list1[i])
                    i += 1
                else:
                    stack.append(list2[j])
                    j += 1

            while i < len1:
                stack.append(list1[i])
                i += 1
            while j < len2:
                stack.append(list2[j])
                j += 1
            
            return stack

        # 获取长度为length 的最大子序列
        def maxSubSeq(seq, length):
            stack = list()
            remain = len(seq) - length

            for num in seq:
                while len(stack) > 0 and stack[-1] < num and remain > 0:
                    stack.pop()
                    remain -= 1
                if len(stack) < length:
                    stack.append(num)
                else:
                    remain -= 1
            return stack


        m, n = len(nums1), len(nums2)
        # 两个数组取的长度分别为x + y = k
        start, end = max(0, k - n), min(k, m)
        res = []
        first = True # 第一次计算
        for x in range(start, end + 1):
            # print(x, k - x)
            digit1 = maxSubSeq(nums1, x)
            digit2 = maxSubSeq(nums2, k - x)

            digit = monoStack(digit1, digit2)
            # print(digit1, digit2, digit)
            if first:
                res = digit
                first = False
            else:
                res = max(digit, res) # 找到字典序最大的

        return res



```class Solution:
    def maxNumber(self, nums1: List[int], nums2: List[int], k: int) -> List[int]:

        # 比较两个剩下的串的字典序和长度大小
        def compare(list1, idx1, list2, idx2):
            len1, len2 = len(list1), len(list2)

            while idx1 < len1 and idx2 < len2:
                d = (list1[idx1] - list2[idx2])
                if d != 0:
                    return d
                idx1 += 1
                idx2 += 1
            
            return (len1 - idx1) - (len2 - idx2)
        # 单调栈
        def monoStack(list1, list2) -> List[int]:
            len1, len2 = len(list1), len(list2)
            stack = list()
            i, j = 0, 0
            while i < len1 and j < len2:
                if compare(list1, i, list2, j) > 0:
                    stack.append(list1[i])
                    i += 1
                else:
                    stack.append(list2[j])
                    j += 1

            while i < len1:
                stack.append(list1[i])
                i += 1
            while j < len2:
                stack.append(list2[j])
                j += 1
            
            return stack

        # 获取长度为length 的最大子序列
        def maxSubSeq(seq, length):
            stack = list()
            remain = len(seq) - length

            for num in seq:
                while len(stack) > 0 and stack[-1] < num and remain > 0:
                    stack.pop()
                    remain -= 1
                if len(stack) < length:
                    stack.append(num)
                else:
                    remain -= 1
            return stack


        m, n = len(nums1), len(nums2)
        # 两个数组取的长度分别为x + y = k
        start, end = max(0, k - n), min(k, m)
        res = []
        first = True # 第一次计算
        for x in range(start, end + 1):
            # print(x, k - x)
            digit1 = maxSubSeq(nums1, x)
            digit2 = maxSubSeq(nums2, k - x)

            digit = monoStack(digit1, digit2)
            # print(digit1, digit2, digit)
            if first:
                res = digit
                first = False
            else:
                res = max(digit, res) # 找到字典序最大的

        return res

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