【leetcode】1253. Reconstruct a 2-Row Binary Matrix

題目以下：

Given the following details of a matrix with n columns and 2 rows :

• The matrix is a binary matrix, which means each element in the matrix can be 0 or 1.
• The sum of elements of the 0-th(upper) row is given as upper.
• The sum of elements of the 1-st(lower) row is given as lower.
• The sum of elements in the i-th column(0-indexed) is colsum[i], where colsum is given as an integer array with length n.

Your task is to reconstruct the matrix with upper, lower and colsum.

Return it as a 2-D integer array.

If there are more than one valid solution, any of them will be accepted.

If no valid solution exists, return an empty 2-D array.

Example 1:

Input: upper = 2, lower = 1, colsum = [1,1,1]
Output: [[1,1,0],[0,0,1]]
Explanation: [[1,0,1],[0,1,0]], and [[0,1,1],[1,0,0]] are also correct answers.

Example 2:

Input: upper = 2, lower = 3, colsum = [2,2,1,1]
Output: []

Example 3:

Input: upper = 5, lower = 5, colsum = [2,1,2,0,1,0,1,2,0,1]
Output: [[1,1,1,0,1,0,0,1,0,0],[1,0,1,0,0,0,1,1,0,1]]

Constraints:

• 1 <= colsum.length <= 10^5
• 0 <= upper, lower <= colsum.length
• 0 <= colsum[i] <= 2

解題思路：colsum[i] = 0 和 colsum[i] = 2的場景很簡單，output[0][i] 和 output[1][i]  都爲0或者都爲1便可。剩下colsum[i] = 1的場景，優先把1分配給output[0][i] ，達到upper上限後，再把剩餘的1分配給output[1][i]。

代碼以下：

class Solution(object):
    def reconstructMatrix(self, upper, lower, colsum):
        """
        :type upper: int
        :type lower: int
        :type colsum: List[int]
        :rtype: List[List[int]]
        """
        res = [[0] * len(colsum) for _ in range(2)]
        one_count = 0
        two_count = 0
        for i in range(len(colsum)):
            if colsum[i] == 2:
                res[0][i] = res[1][i] = 1
                two_count += 1
            elif colsum[i] == 1:one_count += 1
        if upper < two_count or lower < two_count or (upper - two_count + lower - two_count) != one_count:
            return []

        count = upper - two_count
        for i in range(len(colsum)):
            if colsum[i] == 0 or colsum[i] == 2:continue
            if count > 0:
                res[0][i] = 1
                count -= 1
            else:
                res[1][i] = 1

        return res