LeetCode 329. Longest Increasing Path in a Matrix

题意：

Given an integer matrix, find the length of the longest increasing path.

From each cell, you can either move to four directions: left, right, up or down. You may NOT move diagonally or move outside of the boundary (i.e. wrap-around is not allowed).

Example 1:

nums = [
  [9,9,4],
  [6,6,8],
  [2,1,1]
]

 

Return 4
The longest increasing path is [1, 2, 6, 9].

 

思路：

bfs更新。

AC代码：

int path[4][2] = {
    {-1, 0}, {1, 0},{0, -1},{0, 1}
};

class Solution {
public:
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        int n = matrix.size();
        if(n == 0) return 0;
        int m = matrix[0].size();
        vector<int> temp(m, 0);
        vector< vector<int>> dp(n, temp);
        
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<m; j++)
            {
                if(dp[i][j] == 0)
                {
                    queue<int>q;
                    q.push(i*m+j);
                    dp[i][j] = 1;
                    while(!q.empty())
                    {
                        int k = q.front();
                        q.pop();
                        int x = k/m, y = k%m;
                        for(int w = 0; w<4; w++)
                        {
                            int xx = x+path[w][0];
                            int yy = y+path[w][1];
                            if(xx>=0&&xx<n&&yy>=0&&yy<m)
                            {
                                if(matrix[xx][yy] > matrix[x][y] && dp[xx][yy]<dp[x][y]+1)
                                {
                                    dp[xx][yy] = dp[x][y]+1;
                                    q.push(xx*m+yy);
                                }
                            }
                        }
                    }
                }
            }
        }
        int ans = 0;
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<m; j++)
            ans = max(ans, dp[i][j]);
        }
        return ans;
    }
};