[LeetCode] 1253. Reconstruct a 2-Row Binary Matrix 重构 2 行二进制矩阵

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

这道题说是有一个 2 by n 的数组，且只含有0和1两个数字，现在知道第一行的数字和为 upper，第二行的数字和为 lower，且各列的数字和在数组 colsum 中，现在让重建这个数组，若无法重建，则返回空数组。由于原数组中只有0和1，而且只有两行，则每一列的和只有三种情况，0，1，和2。其中0和2的情况最简单，上下两个位置分别只能为0和1，只有当列和为1的时候，才会有不确定性，不知道到底是上面的数字为1还是下面的数字为1。这个时候其实可以使用贪婪算法的思想，判断的依据是此时的 upper 和 lower 的值，当 upper 大于 lower 时，就让上面的数字为1，否则就让下面的数字为1。

这种贪婪算法可以在有解的情况下得到一个合法解，因为这道题没有让返回所有的合法的解。明白了思路，代码就不难写了，遍历每一列，先更新上位数字，若当前列之和为2，则上位数字一定是1，或者列之和为1，且 uppper 大于 lower 的时候，上位数字也是1。再来更新下位数字，若当前列之和为2，则下位数字一定是1，或者列之和为1，且此时上位数字是0的话，则下位数字是1。最后别忘了 upper 和 lower 分别减去当前的上位和下位数字。最后返回的时候判断，若 upper 和 lower 同时为0了，则返回 res，否则返回空数组即可，参见代码如下：

class Solution {
public:
    vector<vector<int>> reconstructMatrix(int upper, int lower, vector<int>& colsum) {
        vector<vector<int>> res(2, vector<int>(colsum.size()));
        for (int i = 0; i < colsum.size(); ++i) {
            res[0][i] = colsum[i] == 2 || (colsum[i] == 1 && upper > lower);
            res[1][i] = colsum[i] == 2 || (colsum[i] == 1 && !res[0][i]);
            upper -= res[0][i];
            lower -= res[1][i];
        }
        return upper == 0 && lower == 0 ? res : vector<vector<int>>();
    }
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1253

类似题目：

Find Valid Matrix Given Row and Column Sums

参考资料：

https://leetcode.com/problems/reconstruct-a-2-row-binary-matrix/

https://leetcode.com/problems/reconstruct-a-2-row-binary-matrix/discuss/425793/C%2B%2BJava-5-lines

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