542. 01 Matrix

最新推荐文章于 2020-10-14 16:17:44 发布

David_DLUT

最新推荐文章于 2020-10-14 16:17:44 发布

阅读量364

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode 文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/David_DLUT/article/details/67655221

LeetCode 专栏收录该内容

42 篇文章

订阅专栏

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:

The number of elements of the given matrix will not exceed 10,000.
There are at least one 0 in the given matrix.
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;
    }
};