leetcode542

来源：互联网发布：不交叉拓扑图算法编辑：程序博客网时间：2024/06/15 01:24

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 00 1 00 0 0

Output:

0 0 00 1 00 0 0

Example 2:
Input:

0 0 00 1 01 1 1

Output:

0 0 00 1 01 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.

这道题我先针对0把0周边的数设为1，后再将改为1的数加入队列中操作

class Solution {public:    vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) {   //找最近的位置和0的距离            vector<vector<int>> empty;if (matrix.empty())return empty;int r = matrix.size();int c = matrix[0].size();vector<vector<int>> result(r, vector<int>(c));queue<int> rq;queue<int> cq;for (int i = 0; i < r; ++i) {for (int j = 0; j < c; ++j) {if (matrix[i][j] == 1) {result[i][j] = INT_MAX;}}}for (int i = 0; i < r; ++i) {for (int j = 0; j < c; ++j) {if (matrix[i][j] == 0) {set(result, i, j, rq, cq);}}}while (!rq.empty() && !cq.empty()) {setcore(result, rq, cq);}return result;    }    void set(vector<vector<int>>& result, int i, int j, queue<int>& rq, queue<int>& cq) {if (i > 0) {if (result[i - 1][j] == INT_MAX) {result[i - 1][j] = 1;rq.push(i - 1);cq.push(j);}}if (i < result.size() - 1) {if (result[i + 1][j] == INT_MAX) {result[i + 1][j] = 1;rq.push(i + 1);cq.push(j);}}if (j > 0) {if (result[i][j - 1] == INT_MAX) {result[i][j - 1] = 1;rq.push(i);cq.push(j - 1);}}if (j < result[0].size() - 1) {if (result[i][j + 1] == INT_MAX) {result[i][j + 1] = 1;rq.push(i);cq.push(j + 1);}}}void setcore(vector<vector<int>>& result, queue<int>& rq, queue<int>& cq) {int i = rq.front();rq.pop();int j = cq.front();cq.pop();if (i > 0) {if (result[i - 1][j] == INT_MAX) {rq.push(i - 1);cq.push(j);}if (result[i - 1][j] == INT_MAX) {result[i - 1][j] = result[i][j]+1;}else if (result[i - 1][j] != 0 && result[i - 1][j] != 1) {result[i - 1][j] = min(result[i][j] + 1, result[i - 1][j]);}}if (i < result.size() - 1) {if ( result[i + 1][j] == INT_MAX) {rq.push(i + 1);cq.push(j);}if (result[i + 1][j] == INT_MAX) {result[i + 1][j] = result[i][j]+1;}else if (result[i + 1][j] != 0 && result[i + 1][j] != 1) {result[i + 1][j] = min(result[i][j] + 1, result[i + 1][j]);}}if (j > 0) {if ( result[i][j - 1]== INT_MAX) {rq.push(i);cq.push(j - 1);}if (result[i][j - 1] == INT_MAX) {result[i][j - 1] = result[i][j]+1;}else if (result[i ][j-1] != 0 && result[i ][j-1] != 1) {result[i][j - 1] = min(result[i][j] + 1, result[i][j - 1]);}}if (j < result[0].size() - 1) {if (result[i][j + 1] == INT_MAX) {rq.push(i);cq.push(j + 1);}if (result[i][j + 1] == INT_MAX) {result[i][j + 1] = result[i][j]+1;}else if (result[i ][j+1] != 0 && result[i ][j+1] != 1) {result[i ][j+1] = min(result[i][j] + 1, result[i][j + 1]);}}}};

代码较混乱，后来想了化简的方法，可以不用分两种进行set，先留下坑，下次填

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