leetcode : Set Matrix Zeroes [技巧 follow up 逐步优化]

最新推荐文章于 2025-09-08 22:40:31 发布

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最新推荐文章于 2025-09-08 22:40:31 发布

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原文链接：http://www.cnblogs.com/superzhaochao/p/6497841.html

本文提供了一种解决矩阵中若某元素为0，则将其所在行与列全置零的问题的方法。采用空间复杂度为O(m+n)的解决方案，并探讨了如何实现常数空间复杂度的算法。

Given a m x 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?

public class Solution {
    public void setZeroes(int[][] matrix) {
        
        		
		if(matrix == null){
			return;
		}
		
		int m = matrix.length;
		int n = matrix[0].length;
		
	    boolean row0 = false;
	    boolean col0 = false;
	    
	    for(int i = 0; i < n; i++){
	        if(matrix[0][i] == 0){
	            row0 = true;
	            break;
	        }
	    }
	    
	    for(int j = 0 ; j < m; j++){
	        if(matrix[j][0] == 0){
	            col0 = true;
	            break;
	        }
	    }
		
        for(int i = 1; i < m; i++){
            for(int j = 1; j < n; j++){
                if(matrix[i][j] == 0){
                    matrix[i][0] = 0;
                    matrix[0][j] = 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(row0 == true){
            for(int i = 0; i < n; i++){
	        matrix[0][i] = 0;
	        }
        }
        
         if(col0 == true){
            for(int i = 0; i < m; i++){
	        matrix[i][0] = 0;
	        }
        }
        
    }
}

转载于:https://www.cnblogs.com/superzhaochao/p/6497841.html