542. 01 Matrix

最新推荐文章于 2020-10-14 16:17:44 发布

转载最新推荐文章于 2020-10-14 16:17:44 发布 · 61 阅读

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原文链接：http://www.cnblogs.com/tsunami-lj/p/6593862.html

本文介绍了一种高效算法，用于计算给定矩阵中每个元素到最近0的距离，通过两次遍历更新距离，确保了计算的准确性。

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.

解题思路：用当前已知的周围（上下左右）最近距离来更新当前节点到0的最近距离，之所以需要遍历两遍是因为第一次假设某些点周围都是1，那就会导致该点的值为无穷大，再倒着遍历的时候他的周边距离变成已知数，所以会更新当前节点的最小值。

class Solution {
public:
    vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) {
        int row=matrix.size(),col=matrix[0].size();
        vector<vector<int> >dp(row,vector<int>(col,INT_MAX));
        for(int times=0;times<=1;times++)
            for(int i=times?row-1:0;times?i>=0:i<row;times?i--:i++)
                for(int j=times?col-1:0;times?j>=0:j<col;times?j--:j++)
                    if(!matrix[i][j]){
                        dp[i][j]=0;
                    }
                    else if(dp[i][j]){
                        if(i&&dp[i-1][j]!=INT_MAX&&dp[i][j]>dp[i-1][j]+1)
                            dp[i][j]=dp[i-1][j]+1;
                        if(j+1<col&&dp[i][j+1]!=INT_MAX&&dp[i][j]>dp[i][j+1]+1)
                            dp[i][j]=dp[i][j+1]+1;
                        if(i+1<row&&dp[i+1][j]!=INT_MAX&&dp[i][j]>dp[i+1][j]+1)
                            dp[i][j]=dp[i+1][j]+1;
                        if(j&&dp[i][j-1]!=INT_MAX&&dp[i][j]>dp[i][j-1]+1)
                            dp[i][j]=dp[i][j-1]+1;
                    }
        return dp;
    }
};

转载于:https://www.cnblogs.com/tsunami-lj/p/6593862.html