leetcode 329. Longest Increasing Path in a Matrix (记忆化DFS剪枝)

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].

Example 2:

nums = [
  [3,4,5],
  [3,2,6],
  [2,2,1]
]

Return 4
The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.

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思路：通过记忆化DFS，剪枝。记忆化DFS主要思想：当搜完某一个节点时，保存这个节点后续的信息，当下次再搜到这个节点时，就不用再次搜索了。leetcode 140.Word Break II比这题难，也可以用记忆化DFS方法形成剪枝。

public int longestIncreasingPath(int[][] matrix) {
        if(matrix==null||matrix.length==0||matrix[0].length==0) return 0;
        int max=0;
        int[][] mem=new int[matrix.length][matrix[0].length];
        for(int i=0;i<matrix.length;i++){
            for(int j=0;j<matrix[0].length;j++){
                max=Math.max(max,dfs(matrix,mem,i,j));
            }
        }
        return max;
    }
    public int dfs(int[][] matrix,int[][] mem,int i,int j){
        if(mem[i][j]!=0){
            return mem[i][j];
        }
        int count=0;
        if(i+1<matrix.length&&matrix[i+1][j]>matrix[i][j]){
            count=Math.max(count,dfs(matrix,mem,i+1,j));
        }
        if(j+1<matrix[0].length&&matrix[i][j+1]>matrix[i][j]){
            count=Math.max(count,dfs(matrix,mem,i,j+1));
        }
        if(i-1>=0&&matrix[i-1][j]>matrix[i][j]){
            count=Math.max(count,dfs(matrix,mem,i-1,j));
        }
        if(j-1>=0&&matrix[i][j-1]>matrix[i][j]){
            count=Math.max(count,dfs(matrix,mem,i,j-1));
        }
        mem[i][j]=count+1;
        return count+1;
    }