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  • 【LeetCode】73. Set Matrix Zeroes (2 solutions)

    Set Matrix Zeroes

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

    click to show follow up.

    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?

    解法一:

    使用数组分别记录需要置零的行列。然后根据数组信息对相应行列置零。

    空间复杂度O(m+n)

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

    解法二:

    使用第一行和第一列记录该行和该列是否应该置零。

    对于由此覆盖掉的原本信息,只要单独遍历第一行第一列判断是否需要置零即可。

    空间复杂度O(1)

    class Solution {
    public:
        void setZeroes(vector<vector<int> > &matrix) {
            if(matrix.empty() || matrix[0].empty())
                return;
            int m = matrix.size();
            int n = matrix[0].size();
            bool col0 = false;
            bool row0 = false;
            for(int i = 0; i < m; i ++)
            {
                if(matrix[i][0] == 0)
                {
                    col0 = true;
                    break;
                }
            }
            for(int j = 0; j < n; j ++)
            {
                if(matrix[0][j] == 0)
                {
                    row0 = 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 ++)
            {
                if(matrix[i][0] == 0)
                {
                    for(int j = 1; j < n; j ++)
                    {
                        matrix[i][j] = 0;
                    }
                }
            }
            for(int j = 1; j < n; j ++)
            {
                if(matrix[0][j] == 0)
                {
                    for(int i = 1; i < m; i ++)
                    {
                        matrix[i][j] = 0;
                    }
                }
            }
            if(col0 == true)
            {
                for(int i = 0; i < m; i ++)
                {
                    matrix[i][0] = 0;
                }
            }
            if(row0 == true)
            {
                for(int j = 0; j < n; j ++)
                {
                    matrix[0][j] = 0;
                }
            }
        }
    };

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  • 原文地址:https://www.cnblogs.com/ganganloveu/p/4153154.html
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