LeetCode Set Matrix Zeroes

LeetCode解题之Set Matrix Zeroes


原题

假设矩阵中存在0。那么把0所在的行和列都置为0。要求在所给的矩阵上完毕操作。

注意点:

  • 最好仅仅申请常量级的额外空间

样例:

输入:

matrix = 
[[1, 0, 1, 1],
[1, 1, 0, 1],
[1, 1, 1, 0],
[1, 1, 1, 1]]

输出:

[[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0],
[1, 0, 0, 0]]

解题思路

一边遍历,一边将对应的行和列置为0是行不通的。会影响后面元素的遍历推断,所以要记录下哪些行和哪些列是要置为0的。为了节约空间,在原矩阵中借两条边,假设该行或者列要置为0,则把左边或者上边的对应位置置为0。假设左边和上边本来就有0,那么须要额外标记一下,最后把左边或者右边也所有置为0.

AC源代码

class Solution(object):
    def setZeroes(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: void Do not return anything, modify matrix in-place instead.
        """
        first_row = False
        first_col = False
        m = len(matrix)
        n = len(matrix[0])
        for i in range(m):
            if matrix[i][0] == 0:
                first_col = True
        for j in range(n):
            if matrix[0][j] == 0:
                first_row = True
        for i in range(1, m):
            for j in range(1, n):
                if matrix[i][j] == 0:
                    matrix[i][0] = matrix[0][j] = 0
        for i in range(1, m):
            for j in range(1, n):
                if matrix[0][j] == 0 or matrix[i][0] == 0:
                    matrix[i][j] = 0
        if first_row:
            for j in range(n):
                matrix[0][j] = 0
        if first_col:
            for i in range(m):
                matrix[i][0] = 0


if __name__ == "__main__":
    matrix = [[1, 0, 1, 1],
              [1, 1, 0, 1],
              [1, 1, 1, 0],
              [1, 1, 1, 1]]
    Solution().setZeroes(matrix)
    assert matrix == [[0, 0, 0, 0],
                      [0, 0, 0, 0],
                      [0, 0, 0, 0],
                      [1, 0, 0, 0]]

欢迎查看我的Github (https://github.com/gavinfish/LeetCode-Python) 来获得相关源代码。

原文地址:https://www.cnblogs.com/jzssuanfa/p/7208413.html