Set Matrix Zeroes

描述

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

Follow up: Did you use extra space?

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?

分析

O(m+n)空间的方法很简单，设置两个 bool 数组，记录每行和每列是否存在 0。

想要常数空间，可以复用第一行和第一列。

代码 1

Java

// Set Matrix Zeroes
// 时间复杂度O(m*n)，空间复杂度O(m+n)
public class Solution {
    public void setZeroes(int[][] matrix) {
        final int m = matrix.length;
        final int n = matrix[0].length;
        boolean[] row = new boolean[m]; // 标记该行是否存在0
        boolean[] col = new boolean[n]; // 标记该列是否存在0

        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (matrix[i][j] == 0) {
                    row[i] = col[j] = true;
                }
            }
        }

        for (int i = 0; i < m; ++i) {
            if (row[i]) Arrays.fill(matrix[i], 0);
        }
        for (int j = 0; j < n; ++j) {
            if (col[j]) {
                for (int i = 0; i < m; ++i) {
                    matrix[i][j] = 0;
                }
            }
        }
    }
}

C++

// Set Matrix Zeroes
// 时间复杂度O(m*n)，空间复杂度O(m+n)
class Solution {
public:
    void setZeroes(vector<vector<int> > &matrix) {
        const size_t m = matrix.size();
        const size_t n = matrix[0].size();
        vector<bool> row(m, false); // 标记该行是否存在0
        vector<bool> col(n, false); // 标记该列是否存在0

        for (size_t i = 0; i < m; ++i) {
            for (size_t j = 0; j < n; ++j) {
                if (matrix[i][j] == 0) {
                    row[i] = col[j] = true;
                }
            }
        }

        for (size_t i = 0; i < m; ++i) {
            if (row[i])
                fill(&matrix[i][0], &matrix[i][0] + n, 0);
        }
        for (size_t j = 0; j < n; ++j) {
            if (col[j]) {
                for (size_t i = 0; i < m; ++i) {
                    matrix[i][j] = 0;
                }
            }
        }
    }
};

代码 2

Java

// Set Matrix Zeroes
// 时间复杂度O(m*n)，空间复杂度O(1)
public class Solution {
    public void setZeroes(int[][] matrix) {
        final int m = matrix.length;
        final int n = matrix[0].length;
        boolean row_has_zero = false; // 第一行是否存在 0
        boolean col_has_zero = false; // 第一列是否存在 0

        for (int i = 0; i < n; i++)
            if (matrix[0][i] == 0) {
                row_has_zero = true;
                break;
            }

        for (int i = 0; i < m; i++)
            if (matrix[i][0] == 0) {
                col_has_zero = true;
                break;
            }

        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[i][0] == 0 || matrix[0][j] == 0)
                    matrix[i][j] = 0;
        if (row_has_zero)
            for (int i = 0; i < n; i++)
                matrix[0][i] = 0;
        if (col_has_zero)
            for (int i = 0; i < m; i++)
                matrix[i][0] = 0;
    }
};

C++

// Set Matrix Zeroes
// 时间复杂度O(m*n)，空间复杂度O(1)
class Solution {
public:
    void setZeroes(vector<vector<int> > &matrix) {
        const size_t m = matrix.size();
        const size_t n = matrix[0].size();
        bool row_has_zero = false; // 第一行是否存在 0
        bool col_has_zero = false; // 第一列是否存在 0

        for (size_t i = 0; i < n; i++)
            if (matrix[0][i] == 0) {
                row_has_zero = true;
                break;
            }

        for (size_t i = 0; i < m; i++)
            if (matrix[i][0] == 0) {
                col_has_zero = true;
                break;
            }

        for (size_t i = 1; i < m; i++)
            for (size_t j = 1; j < n; j++)
                if (matrix[i][j] == 0) {
                    matrix[0][j] = 0;
                    matrix[i][0] = 0;
                }
        for (size_t i = 1; i < m; i++)
            for (size_t j = 1; j < n; j++)
                if (matrix[i][0] == 0 || matrix[0][j] == 0)
                    matrix[i][j] = 0;
        if (row_has_zero)
            for (size_t i = 0; i < n; i++)
                matrix[0][i] = 0;
        if (col_has_zero)
            for (size_t i = 0; i < m; i++)
                matrix[i][0] = 0;
    }
};