542. 01 Matrix

Medium

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Given a matrix consists of 0 and 1, find the distance of the nearest 0 for each cell.

The distance between two adjacent cells is 1.

Example 1:

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

Output:
[[0,0,0],
 [0,1,0],
 [0,0,0]]

Example 2:

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

Output:
[[0,0,0],
 [0,1,0],
 [1,2,1]]

Note:

1. The number of elements of the given matrix will not exceed 10,000.
2. There are at least one 0 in the given matrix.
3. The cells are adjacent in only four directions: up, down, left and right.

Accepted

46,846

Submissions

129,726

 

 

思路:

1 如果格子内元素本身是0, 不需要计算, 自己到自己的距离为0

2 如果格子内元素非0, 自然想到DP, 上下左右四个方向的元素,找出最小的, 再加1

3 由于格子是二维矩阵, 然而代码遍历是一维遍历. 所以需要遍历两遍. 左上+右下 或者  右上+左下  都可以.  想象一下 要是三维矩阵... 估计得遍历四次了... 再follow up N维矩阵怎么办...

 

class Solution {
    public int[][] updateMatrix(int[][] matrix) {
        if(null==matrix||matrix.length<1)return null;
        
        int width=matrix[0].length;
        int height=matrix.length;
        int [][]res=new int[height][width];
        for(int j=0;j<height;++j)
            for(int i=0;i<width;++i)
                res[j][i]=Integer.MAX_VALUE-1;  //这里有个坑点,如果不-1, 得到完全错误的结果. 因为底下的dp会+1, 就溢出变成负数....; 要么这里选择稍小的数字
        
        for(int j=0;j<height;++j)
            for(int i=0;i<width;++i)
            {
                if(0==matrix[j][i])
                    res[j][i]=0;
                else
                {
                    if(i>0)
                        res[j][i]=Math.min(res[j][i], res[j][i-1]+1);
                    if(j>0)
                        res[j][i]=Math.min(res[j][i], res[j-1][i]+1);
                }
            }
        
        for(int j=height-1;j>=0;--j)
            for(int i=width-1;i>=0;--i)
            {
                if(0==matrix[j][i])res[j][i]=0;
                else
                {
                    if(j<height-1)
                        res[j][i]=Math.min(res[j][i], res[j+1][i]+1);
                    if(i<width-1)
                        res[j][i]=Math.min(res[j][i], res[j][i+1]+1);
                }
            }
        return res;
        
    }
}