一、题目描述

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.

二、题解

暴力的思路很简单：初始化一个dists数组记录0与1的距离

• 当<i,j>位置为0时，dists数组自然记录该位置为0
• 当<i,j>位置为1时，从头开始遍历，在遍历的过程中只要位置<k,l>是0，计算与<i,j>位置的距离dist_01，在这个过程中取最小的dist_01

(1)暴力破解

public class Main {
  public static int[][] updateMatrix(int[][] matrix) {
    int rows = matrix.length;
    int cols = matrix[0].length;
    int max=Integer.MAX_VALUE;
    int[][] dists = new int[rows][cols];
    for (int i = 0; i < rows; i++) {
      for (int j = 0; j < cols; j++) {
        dists[i][j]=max;
        if(matrix[i][j]==0)
          dists[i][j]=0;
        else if(matrix[i][j]==1) {
          for (int k = 0; k < rows; k++)
          for (int l = 0; l < cols; l++)
          if(matrix[k][l]==0) {
            int dist_01 = Math.abs(i-k)+Math.abs(j-l);
            dists[i][j]=Math.min(dists[i][j], dist_01);
          }
        }
      }
    }
    return dists;
  }
  public static void main(String[] args) {
    int[][] m = new int[][] {
      {0,0,0},
      {0,1,0},
      {0,0,0}
    };
    updateMatrix(m);
  }
}

复杂度分析

• 时间复杂度：O((r⋅c)^2) )，对于每个为 1 的点都需要遍历整个矩阵一次。
• 空间复杂度：O(r⋅c)，dists数组的开销