329. Longest Increasing Path in a Matrix

Posted 2020-11-30 wentiliangkaihua

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:

Input: nums = 
[
  [9,9,4],
  [6,6,8],
  [2,1,1]
] 
Output: 4 
Explanation: The longest increasing path is [1, 2, 6, 9].

Example 2:

Input: nums = 
[
  [3,4,5],
  [3,2,6],
  [2,2,1]
] 
Output: 4 
Explanation: The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.

public class Solution {
    public int longestIncreasingPath(int[][] matrix) {
        if(matrix==null||matrix.length==0) return 0;
        int row=matrix.length,col=matrix[0].length;
        int result[][]=new int[row][col]; //store the longest path at every node
        int max=1;
        for(int[]arr:result){
            Arrays.fill(arr,-1);
        }
        for(int i=0;i<row;i++){
            for(int j=0;j<col;j++){
                findPath(matrix,result,i,j);
                max=Math.max(result[i][j],max);
            }
        }
        return max;
    }
    
    private void findPath(int[][]matrix,int result[][],int i,int j){
        int row=matrix.length,col=matrix[0].length;
        
        int xPos[]={0,0,1,-1};
        int yPos[]={1,-1,0,0};
        int max=1;
        for(int k=0;k<4;k++){  //at four direction
            
                int nX=xPos[k]+i;
                int nY=yPos[k]+j;
                if(nX>=row||nX<0||nY>=col||nY<0){ //skip when out of boundary
                    continue;
                }
                if(matrix[nX][nY]>matrix[i][j]){
                    if(result[nX][nY]<0){  //haven‘t visited [nX,nY]
                        findPath(matrix,result,nX,nY);
                    }
                    max=Math.max( 1+result[nX][nY],max);
                }
            
        }
       result[i][j]=max;
        
    }
}

答案一：用数组表示当前元素拥有的最大增长path，上下左右四个方向用0，±1表示，很巧妙，然后是dfs

class Solution {
    public int longestIncreasingPath(int[][] matrix) {
        int max = 0;
        if(matrix.length == 0)
            return 0;
        int[][] holder = new int[matrix.length][matrix[0].length];
        for(int i = 0; i < matrix.length; i++){
            for(int j = 0; j < matrix[0].length; j++){
                int _max = pathFinder(matrix, holder, Integer.MIN_VALUE, i, j);
                max = _max > max ? _max : max;
            }
        }
        return max;
    }
    
    
    private int pathFinder(int[][] matrix, int[][] dp, int last, int v, int w){
        if(v < 0 || w < 0 || v == matrix.length || w == matrix[0].length || matrix[v][w] <= last)
            return 0;
        if(dp[v][w] > 0){
            return dp[v][w];
        }
        int val = pathFinder(matrix, dp, matrix[v][w], v-1, w);
        val = Math.max(pathFinder(matrix, dp, matrix[v][w], v+1, w), val);
        val = Math.max(pathFinder(matrix, dp, matrix[v][w], v, w-1), val);
        val = Math.max(pathFinder(matrix, dp, matrix[v][w], v, w+1), val);
        dp[v][w] = ++val;
        return val;
    }
}

答案2更厉害了，但是本质和1一样