[LeetCode] 1253. Reconstruct a 2-Row Binary Matrix 重構 2 行二進制矩陣

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

這道題說是有一個 2 by n 的數組，且只含有0和1兩個數字，現在知道第一行的數字和為 upper，第二行的數字和為 lower，且各列的數字和在數組 colsum 中，現在讓重建這個數組，若無法重建，則返回空數組。由於原數組中只有0和1，而且只有兩行，則每一列的和只有三種情況，0，1，和2。其中0和2的情況最簡單，上下兩個位置分別只能為0和1，只有當列和為1的時候，才會有不確定性，不知道到底是上面的數字為1還是下面的數字為1。這個時候其實可以使用貪婪算法的思想，判斷的依據是此時的 upper 和 lower 的值，當 upper 大於 lower 時，就讓上面的數字為1，否則就讓下面的數字為1。

這種貪婪算法可以在有解的情況下得到一個合法解，因為這道題沒有讓返回所有的合法的解。明白了思路，代碼就不難寫了，遍歷每一列，先更新上位數字，若當前列之和為2，則上位數字一定是1，或者列之和為1，且 uppper 大於 lower 的時候，上位數字也是1。再來更新下位數字，若當前列之和為2，則下位數字一定是1，或者列之和為1，且此時上位數字是0的話，則下位數字是1。最后別忘了 upper 和 lower 分別減去當前的上位和下位數字。最后返回的時候判斷，若 upper 和 lower 同時為0了，則返回 res，否則返回空數組即可，參見代碼如下：

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

類似題目：

Find Valid Matrix Given Row and Column Sums

參考資料：

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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