1253. Reconstruct a 2-Row Binary Matrix

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]]

class Solution {
    List<List<Integer>> res;
    public List<List<Integer>> reconstructMatrix(int upper, int lower, int[] colsum) {
        res = new ArrayList<>();
        int n = colsum.length;
        List<Integer> first = new ArrayList<>();
        List<Integer> second = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            if (colsum[i] == 2) {
                first.add(1);
                second.add(1);
                upper--;
                lower--;
            } else if (colsum[i] == 0) {
                first.add(0);
                second.add(0); 
            } else {
                if (upper > lower) {
                    first.add(1);
                    upper--;
                    second.add(0);
                } else {
                    second.add(1);
                    lower--;
                    first.add(0);
                }
            }
            if (upper < 0 || lower < 0) {
                return res;
            }
        }
        if (lower == 0 && upper == 0) {
            res.add(first);
            res.add(second);
        }
        return res;
    }
}

简化版的：

class Solution {
    public List<List<Integer>> reconstructMatrix(int upper, int lower, int[] colsum) {
        int[][] res = new int[2][colsum.length];
        for(int i = 0; i < colsum.length; i++){
            res[0][i] = (colsum[i] == 2 || (colsum[i] == 1 && lower < upper)) ? 1 : 0; 
            res[1][i] = (colsum[i] == 2 || (colsum[i] == 1 && res[0][i] != 1)) ? 1 : 0; 
            upper -= (res[0][i] == 1) ? 1 : 0;
            lower -= (res[1][i] == 1) ? 1 : 0;
        }
        return lower == 0 && upper == 0 ? new ArrayList(Arrays.asList(res[0], res[1])) : new ArrayList();  
    }
}

意思是重造数组，upper和lower表示行的sum，colsum数组顾名思义

所以colsum的值等于2时两个数组都为1，如果等于1则让lower或upper中大的一个为1，因为有且只有一种可能。

最后判断upper和lower是不是等于0，如果是就成功，否则失败。