這道題我用最容易想的想法就是回溯法計算分別延四個方向計算，結果時間複雜度為O(n5),結果導致時間過長

程式碼如下

public class Solution {
    int max = 0;
    public int longestIncreasingPath(int[][] matrix) {
        for(int i = 0;i <matrix.length;i++){
            for(int j = 0;j<matrix[i].length;j++){
                go(matrix,i,j,0);
            }
        }
        return max;
    }
    public void go(int[][] matrix,int i,int j,int pathNum){
        pathNum++;
        //System.out.println(pathNum);
        if(i-1>=0&&matrix[i-1][j] > matrix[i][j])
            go(matrix,i-1,j,pathNum);
        if(i+1<matrix.length&&matrix[i+1][j] > matrix[i][j])
            go(matrix,i+1,j,pathNum);
        if(matrix.length>0 && j-1>=0 &&matrix[i][j-1] > matrix[i][j])
            go(matrix,i,j-1,pathNum);
        if(matrix.length>0 && j+1< matrix[0].length&&matrix[i][j+1]>matrix[i][j])
            go(matrix,i,j+1,pathNum);
        
        max = Math.max(pathNum,max);
    }
 

if四個方向由比matrix[i][j]大的則dp[i][j] = max(四個方向比他大的+1);

else dp[i][j] = 1

時間複雜度為O(5n)

程式碼如下

public class Solution {
    
    int[][] dp;
    public int longestIncreasingPath(int[][] matrix) {
        if(matrix.length == 0)
            return 0;
        int max = 0;
        dp = new int[matrix.length][matrix[0].length];
        for(int i = 0;i <matrix.length;i++){
            for(int j = 0;j<matrix[i].length;j++){
                max = Math.max(go(matrix,i,j),max);
            }
        }
        return max;
    }
    public int go(int[][] matrix,int i,int j){
        if(dp[i][j]!=0)
            return dp[i][j];
        dp[i][j] = 1;
        if(i-1>=0&&matrix[i-1][j] > matrix[i][j])
            dp[i][j] = Math.max(1+go(matrix,i-1,j),dp[i][j]);
        if(i+1<matrix.length&&matrix[i+1][j] > matrix[i][j])
            dp[i][j] = Math.max(1+go(matrix,i+1,j),dp[i][j]);
        if(matrix.length>0 && j-1>=0 &&matrix[i][j-1] > matrix[i][j])
            dp[i][j] = Math.max(1+go(matrix,i,j-1),dp[i][j]);
        if(matrix.length>0 && j+1< matrix[0].length&&matrix[i][j+1]>matrix[i][j])
            dp[i][j] = Math.max(1+go(matrix,i,j+1),dp[i][j]);
        
        return dp[i][j];
    }
}