LeetCode 1253. Reconstruct a 2-Row Binary Matrix

1、原题描述

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

2、简要翻译：

构建一个两行多列的数据，第一行的和为upper，第二行的和为lower，第i 列的和为 colsum[i] 。 如果不存在这样的数组返回空值

3、代码分析：

• 如果upper + lower 不等于 colsum的和，那么返回空。
• 如果 colsum 中2的个数大于upper或者 lower，返回空
• 从左到右，尽量保持上下均衡的前提下中插入1。（当colums[i]==1的时候，如果lower >= upper 插下，其余情况插上）

4、解答代码：

public List<List<Integer>> reconstructMatrix(int upper, int lower, int[] colums) {
        int sum = 0;
        int numof2 = 0;
        for (int colum : colums) {
            sum += colum;
            if (colum == 2) {
                numof2++;
            }
        }
        List<List<Integer>> result = new ArrayList<>();
        if (sum != upper + lower || numof2 > Math.min(upper, lower)) {
            return result;
        }
        List<Integer> upperList = new ArrayList<>();
        List<Integer> lowerList = new ArrayList<>();
        result.add(upperList);
        result.add(lowerList);
        for (int colum : colums) {
            if (colum == 0) {
                upperList.add(0);
                lowerList.add(0);
            } else if (colum == 2) {
                upperList.add(1);
                upper--;
                lowerList.add(1);
                lower--;
            } else {
                if (upper > lower) {
                    upperList.add(1);
                    lowerList.add(0);
                    upper--;
                } else {
                    upperList.add(0);
                    lowerList.add(1);
                    lower--;
                }
            }
        }
        return result;
    }