mycode

空间复杂度 m+n

思路:用set把为0元素所在行、列记录下来

注意:注释的方法更快

class Solution(object):
    def setZeroes(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: None Do not return anything, modify matrix in-place instead.
        """
        rows = set()
        cols = set()
        count_row = len(matrix)
        count_col = len(matrix[0])
        for i in range(count_row):
            for j in range(count_col):
                if matrix[i][j] == 0:
                    rows.add(i)
                    cols.add(j)
        #for row in rows:
        #    matrix[row] = [0]*count_col
        #for col in cols:
        #    for line in matrix:
        #        line[col] = 0
        for i in rows:
            for j in xrange(count_col):
                matrix[i][j] = 0
        for i in xrange(count_row):
            for j in cols:
                matrix[i][j] = 0
        return matrix

参考:

常数空间,用第一行第一列来记录

class Solution(object):
    def setZeroes(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: None Do not return anything, modify matrix in-place instead.
        """
        first_row = False
        first_col = False
        m = len(matrix)
        n = len(matrix[0]) #[[]]
        if m == 0 or n == 0 : 
            return
        flag = -1
        for i in range(m):
            if matrix[i][0] == 0: #说明第一列本来最后就应该都等于0,以免被覆盖
                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
        #第一行所有列,第一列所有行 为对照表,所以节省了空间
        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