542. 01 Matrix

本文介绍了一种使用广度优先搜索算法来解决矩阵中每个元素到最近的0元素距离的问题。通过遍历矩阵,将所有0元素作为起始点进行广度优先搜索,逐步更新相邻元素的距离值。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

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. 

思想:广度优先遍历

class Solution {
public:
    vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) {
        int m = matrix.size();
        int n = matrix[0].size();
        vector<vector<int>> res(m, vector<int>(n, -1));
        vector<int> index_i(m * n, -1);
        vector<int> index_j(m * n, -1);
        int beg = 0;
        int last = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (matrix[i][j] == 0) {
                    res[i][j] = 0;
                    index_i[last] = i;
                    index_j[last] = j;
                    last++;
                }
            }
        }
        last = last % (m * n);
        while(beg != last) {
            int cur_i = index_i[beg];
            int cur_j = index_j[beg];
            if ((cur_i > 0) && (res[cur_i-1][cur_j] == -1)) {
                res[cur_i-1][cur_j] = res[cur_i][cur_j] + 1;
                index_i[last] = cur_i - 1;
                index_j[last] = cur_j;
                last = (last + 1) % (m * n);
            }
            if ((cur_i < m - 1) && (res[cur_i+1][cur_j] == -1)) {
                res[cur_i+1][cur_j] = res[cur_i][cur_j] + 1;
                index_i[last] = cur_i + 1;
                index_j[last] = cur_j;
                last = (last + 1) % (m * n);
            }
            if ((cur_j > 0) && (res[cur_i][cur_j-1] == -1)) {
                res[cur_i][cur_j-1] = res[cur_i][cur_j] + 1;
                index_i[last] = cur_i;
                index_j[last] = cur_j - 1;
                last = (last + 1) % (m * n);
            }
            if ((cur_j < n - 1) && (res[cur_i][cur_j+1] == -1)) {
                res[cur_i][cur_j+1] = res[cur_i][cur_j] + 1;
                index_i[last] = cur_i;
                index_j[last] = cur_j + 1;
                last = (last + 1) % (m * n);
            }
            beg = (beg + 1) % (m * n);
        }
        return res;
    }
};


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值