题目

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.

给定一个由0和1组成的矩阵，找到每个单元和最近的0的距离。

两个相邻单元的距离是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.

代码

#include <vector>
#include <iostream>  
using namespace std;

class Solution {
public:
	bool valid(vector<vector<int>> & matrix, int m, int n, int i, int j, int val) {
		return i >= 0 && i < m&&j >= 0 && j < n&&matrix[i][j] == val;
	}

	vector<vector<int>> updateMatrix(vector<vector<int>> & matrix) {
		int rows = matrix.size();
		if (rows == 0) {
			return matrix;
		}
		int cols = matrix[0].size();

		vector<vector<int>> dist(rows, vector<int>(cols, INT_MAX));
		for (int i = 0; i<rows; i++) {
			for (int j = 0; j<cols; j++) {
				dist[i][j] = matrix[i][j];
			}
		}
		int val = 0;
		bool idx = true;
		while (idx) {
			int next = val + 1;
			int last = next + 1;
			idx = false;
			for (int i = 0; i<rows; i++) {
				for (int j = 0; j<cols; j++) {
					if (dist[i][j] >= next) {
						if (valid(dist, rows, cols, i - 1, j, val) || valid(dist, rows, cols, i + 1, j, val) || valid(dist, rows, cols, i, j - 1, val) || valid(dist, rows, cols, i, j + 1, val)) {
							dist[i][j] = next;

						}
						else {
							dist[i][j] = last;
						}
						idx = true;
					}
				}
			}
			val = val + 1;
			for (int i = 0; i < 5; i++) {
				for (int j = 0; j < 5; j++) {
					cout << dist[i][j] << '\t';
				}
				cout << endl;
			}
		}
		return dist;
	}
};

int main(){
	vector<vector<int>> src(5, vector<int>(5,1));
	src[2][2] = 0;
	Solution so;
	vector<vector<int>> dist = so.updateMatrix(src);
	for (int i = 0; i < 5; i++) {
		for (int j = 0; j < 5; j++) {
			cout << dist[i][j] << '\t';
		}
		cout << endl;
	}
}

结果