[LeetCode] 73. Set Matrix Zeroes 矩阵赋零

Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in-place.

Example 1:

Input:
[
  [1,1,1],
  [1,0,1],
  [1,1,1]
]
Output:
[
  [1,0,1],
  [0,0,0],
  [1,0,1]
]

Example 2:

Input:
[
  [0,1,2,0],
  [3,4,5,2],
  [1,3,1,5]
]
Output:
[
  [0,0,0,0],
  [0,4,5,0],
  [0,3,1,0]
]

Follow up:

• A straight forward solution using O(mn) space is probably a bad idea.
• A simple improvement uses O(m + n) space, but still not the best solution.
• Could you devise a constant space solution?

 

给一个m x n的矩阵，如果一个元素是0，就把它所在的行和列都设成0，用in place做。

解法1: 新建一个矩阵，然后一行一行的扫，只要有0，就将新建的矩阵的对应行全赋0，行扫完再扫列，然后把更新完的矩阵赋给matrix。空间复杂度为O(mn)。

解法2: 用一个长度为m的一维数组记录各行中是否有0，用一个长度为n的一维数组记录各列中是否有0，最后直接更新matrix数组即可。空间复杂度为O(m+n)，

解法3: 这道题要求用常数级空间复杂度O(1)，不能新建数组，就用原数组的第一行第一列来记录各行各列是否有0.

- 先扫描第一行第一列，如果有0，则将各自的flag设置为true
- 然后扫描除去第一行第一列的整个数组，如果有0，则将对应的第一行和第一列的数字赋0
- 再次遍历除去第一行第一列的整个数组，如果对应的第一行和第一列的数字有一个为0，则将当前值赋0
- 最后根据第一行第一列的flag来更新第一行第一列

时间复杂度是O(m*n), 三种方法都一样，需要进行两次扫描，一次确定行列置0情况，一次对矩阵进行实际的置0操作。

Java:

void setZeroes(vector<vector<int> > &matrix) {
    int col0 = 1, rows = matrix.size(), cols = matrix[0].size();

    for (int i = 0; i < rows; i++) {
        if (matrix[i][0] == 0) col0 = 0;
        for (int j = 1; j < cols; j++)
            if (matrix[i][j] == 0)
                matrix[i][0] = matrix[0][j] = 0;
    }

    for (int i = rows - 1; i >= 0; i--) {
        for (int j = cols - 1; j >= 1; j--)
            if (matrix[i][0] == 0 || matrix[0][j] == 0)
                matrix[i][j] = 0;
        if (col0 == 0) matrix[i][0] = 0;
    }
}　　

Java:

public void setZeroes(int[][] matrix) {
    boolean fr = false,fc = false;
    for(int i = 0; i < matrix.length; i++) {
        for(int j = 0; j < matrix[0].length; j++) {
            if(matrix[i][j] == 0) {
                if(i == 0) fr = true;
                if(j == 0) fc = true;
                matrix[0][j] = 0;
                matrix[i][0] = 0;
            }
        }
    }
    for(int i = 1; i < matrix.length; i++) {
        for(int j = 1; j < matrix[0].length; j++) {
            if(matrix[i][0] == 0 || matrix[0][j] == 0) {
                matrix[i][j] = 0;
            }
        }
    }
    if(fr) {
        for(int j = 0; j < matrix[0].length; j++) {
            matrix[0][j] = 0;
        }
    }
    if(fc) {
        for(int i = 0; i < matrix.length; i++) {
            matrix[i][0] = 0;
        }
    }

}　　

Java:

public class Solution {
    public void setZeroes(int[][] matrix) {
        boolean firstRowZero = false;
        boolean firstColumnZero = false;

        //set first row and column zero or not
        for(int i=0; i<matrix.length; i++){
            if(matrix[i][0] == 0){
                firstColumnZero = true;
                break;
            }
        }

        for(int i=0; i<matrix[0].length; i++){
            if(matrix[0][i] == 0){
                firstRowZero = true;
                break;
            }
        }

        //mark zeros on first row and column
        for(int i=1; i<matrix.length; i++){
            for(int j=1; j<matrix[0].length; j++){
                if(matrix[i][j] == 0){
                   matrix[i][0] = 0;
                   matrix[0][j] = 0;
                }
            }
        }

        //use mark to set elements
        for(int i=1; i<matrix.length; i++){
            for(int j=1; j<matrix[0].length; j++){
                if(matrix[i][0] == 0 || matrix[0][j] == 0){
                   matrix[i][j] = 0;
                }
            }
        }

        //set first column and row
        if(firstColumnZero){
            for(int i=0; i<matrix.length; i++)
                matrix[i][0] = 0;
        }

        if(firstRowZero){
            for(int i=0; i<matrix[0].length; i++)
                matrix[0][i] = 0;
        }

    }
}　　

Java:

public void setZeroes(int[][] matrix) {
    if(matrix==null || matrix.length==0 || matrix[0].length==0)
        return;
    boolean rowFlag = false;
    boolean colFlag = false;
    for(int i=0;i<matrix.length;i++)
    {
        if(matrix[i][0]==0)
        {
            colFlag = true;
            break;
        }
    }
    for(int i=0;i<matrix[0].length;i++)
    {
        if(matrix[0][i]==0)
        {
            rowFlag = true;
            break;
        }
    }
    for(int i=1;i<matrix.length;i++)
    {
        for(int j=1;j<matrix[0].length;j++)
        {
            if(matrix[i][j]==0)
            {
                matrix[i][0] = 0;
                matrix[0][j] = 0;
            }
        }
    }
    for(int i=1;i<matrix.length;i++)
    {
        for(int j=1;j<matrix[0].length;j++)
        {
            if(matrix[i][0]==0 || matrix[0][j]==0)
                matrix[i][j] = 0;
        }
    }
    if(colFlag)
    {
        for(int i=0;i<matrix.length;i++)
        {
            matrix[i][0] = 0;
        }
    }
    if(rowFlag)
    {
        for(int i=0;i<matrix[0].length;i++)
        {
            matrix[0][i] = 0;
        }
    }
}  　　

Python:

class Solution:
    # @param matrix, a list of lists of integers
    # RETURN NOTHING, MODIFY matrix IN PLACE.
    def setZeroes(self, matrix):
        first_col = reduce(lambda acc, i: acc or matrix[i][0] == 0, xrange(len(matrix)), False)
        first_row = reduce(lambda acc, j: acc or matrix[0][j] == 0, xrange(len(matrix[0])), False)

        for i in xrange(1, len(matrix)):
            for j in xrange(1, len(matrix[0])):
                if matrix[i][j] == 0:
                    matrix[i][0], matrix[0][j] = 0, 0

        for i in xrange(1, len(matrix)):
            for j in xrange(1, len(matrix[0])):
                if matrix[i][0] == 0 or matrix[0][j] == 0:
                    matrix[i][j] = 0

        if first_col:
            for i in xrange(len(matrix)):
                matrix[i][0] = 0

        if first_row:
            for j in xrange(len(matrix[0])):
                matrix[0][j] = 0　　

Python:

class Solution:
# @param {integer[][]} matrix
# @return {void} Do not return anything, modify matrix in-place instead.
def setZeroes(self, matrix):
    m = len(matrix)
    if m == 0:
        return
    n = len(matrix[0])

    row_zero = False
    for i in range(m):
        if matrix[i][0] == 0:
            row_zero = True
    col_zero = False
    for j in range(n):
        if matrix[0][j] == 0:
            col_zero = True

    for i in range(1, m):
        for j in range(1, n):
            if matrix[i][j] == 0:
                matrix[i][0] = 0
                matrix[0][j] = 0

    for i in range(1, m):
        if matrix[i][0] == 0:
            for j in range(1, n):
                matrix[i][j] = 0

    for j in range(1, n):
        if matrix[0][j] == 0:
            for i in range(1, m):
                matrix[i][j] = 0

    if col_zero:
        for j in range(n):
            matrix[0][j] = 0
    if row_zero:
        for i in range(m):
            matrix[i][0] = 0

C++:

class Solution {
public:
    void setZeroes(vector<vector<int> > &matrix) {
        if (matrix.empty() || matrix[0].empty()) return;
        int m = matrix.size(), n = matrix[0].size();
        bool rowZero = false, colZero = false;
        for (int i = 0; i < m; ++i) {
            if (matrix[i][0] == 0) colZero = true;
        }
        for (int i = 0; i < n; ++i) {
            if (matrix[0][i] == 0) rowZero = true;
        }
        for (int i = 1; i < m; ++i) {
            for (int j = 1; j < n; ++j) {
                if (matrix[i][j] == 0) {
                    matrix[0][j] = 0;
                    matrix[i][0] = 0;
                }
            }
        }
        for (int i = 1; i < m; ++i) {
            for (int j = 1; j < n; ++j) {
                if (matrix[0][j] == 0 || matrix[i][0] == 0) {
                    matrix[i][j] = 0;
                }
            }
        }
        if (rowZero) {
            for (int i = 0; i < n; ++i) matrix[0][i] = 0;
        }
        if (colZero) {
            for (int i = 0; i < m; ++i) matrix[i][0] = 0;
        }
    }
};

　　

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