[LeetCode]Longest Increasing Path in a Matrix(Java)

来源：互联网发布：java开发薪资编辑：程序博客网时间：2024/06/04 19:16

这道题我用最容易想的想法就是回溯法计算分别延四个方向计算，结果时间复杂度为O(n5),结果导致时间过长

代码如下

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);    }

后期看提示改用动态规划做，即

if四个方向由比matrix[i][j]大的则dp[i][j] = max(四个方向比他大的+1);

else dp[i][j] = 1

时间复杂度为O(5n)

代码如下

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];    }}

2017/3/5

0 0