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< p > Given an < code > m x n< / code > integer matrix < code > matrix< / code > , if an element is < code > 0< / code > , set its entire row and column to < code > 0< / code > ' s.< / p >
< p > You must do it < a href = "https://en.wikipedia.org/wiki/In-place_algorithm" target = "_blank" > in place< / a > .< / p >
< p > < / p >
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< p > < strong class = "example" > Example 1:< / strong > < / p >
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< img alt = "" src = "https://assets.leetcode.com/uploads/2020/08/17/mat1.jpg" style = "width: 450px; height: 169px;" / >
< pre >
< strong > Input:< / strong > matrix = [[1,1,1],[1,0,1],[1,1,1]]
< strong > Output:< / strong > [[1,0,1],[0,0,0],[1,0,1]]
< / pre >
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< p > < strong class = "example" > Example 2:< / strong > < / p >
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< img alt = "" src = "https://assets.leetcode.com/uploads/2020/08/17/mat2.jpg" style = "width: 450px; height: 137px;" / >
< pre >
< strong > Input:< / strong > matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]]
< strong > Output:< / strong > [[0,0,0,0],[0,4,5,0],[0,3,1,0]]
< / pre >
< p > < / p >
< p > < strong > Constraints:< / strong > < / p >
< ul >
< li > < code > m == matrix.length< / code > < / li >
< li > < code > n == matrix[0].length< / code > < / li >
< li > < code > 1 < = m, n < = 200< / code > < / li >
< li > < code > -2< sup > 31< / sup > < = matrix[i][j] < = 2< sup > 31< / sup > - 1< / code > < / li >
< / ul >
< p > < / p >
< p > < strong > Follow up:< / strong > < / p >
< ul >
< li > A straightforward solution using < code > O(mn)< / code > space is probably a bad idea.< / li >
< li > A simple improvement uses < code > O(m + n)< / code > space, but still not the best solution.< / li >
< li > Could you devise a constant space solution?< / li >
< / ul >
