leetcode 329 : Longest Increasing Path in a Matrix : dfs+dp

329. Longest Increasing Path in a Matrix

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Question

Total Accepted: 9010  Total Submissions: 29883  Difficulty: Hard

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.

这道题目我使用的dfs+dp，这里dp的实现的方式是使用hashmap存储已经计算过的的节点，然后dfs搜索到已经计算过的结果，直接提取结果。

public class Solution {
    HashMap<Integer, Integer> index2lenth=new HashMap<>();
	public int Search(int iInex,int jIndex,int[][]matrix){
		int row=matrix.length;
		int column=matrix[0].length;
		if(index2lenth.containsKey(iInex*column+jIndex))
			return index2lenth.get(iInex*column+jIndex);
		int max=1;
		int temp;
		if(iInex-1>=0&&matrix[iInex-1][jIndex]<matrix[iInex][jIndex]){
			temp=Search(iInex-1, jIndex, matrix);
			max=max<(temp+1)?temp+1:max;
		}
		if(jIndex-1>=0&&matrix[iInex][jIndex-1]<matrix[iInex][jIndex]){
			temp=Search(iInex, jIndex-1, matrix);
			max=max<(temp+1)?temp+1:max;
		}
		if(iInex+1<row&&matrix[iInex+1][jIndex]<matrix[iInex][jIndex]){
			temp=Search(iInex+1, jIndex, matrix);
			max=max<(temp+1)?temp+1:max;
		}
		if(jIndex+1<column&&matrix[iInex][jIndex+1]<matrix[iInex][jIndex]){
			temp=Search(iInex, jIndex+1, matrix);
			max=max<(temp+1)?temp+1:max;
		}
		index2lenth.put(iInex*column+jIndex, max);
		return max;
	}
	public int longestIncreasingPath(int[][] matrix) {
		if(matrix.length==0)
			return 0;
		
		int max=0;
		int row=matrix.length;
		int column=matrix[0].length;
		
		for(int i=0;i<row;i++){
			for(int j=0;j<column;j++){
				if(index2lenth.containsKey(i*column+j)){
					int temp=index2lenth.get(i*column+j);
					max=max<temp?temp:max;
				}
				else {
					int temp=Search(i,j,matrix);
					max=max<temp?temp:max;
				}
			}
		}
		return max;
	}
}

总觉得有纯dp的方式，尝试了一下，超时了，求指导~

public class Solution {
    public int longestIncreasingPath(int[][] matrix) {
		if(matrix.length==0)
			return 0;
		int rowNum=matrix.length;
		int column=matrix[0].length;
		int[][] lenth=new int[rowNum][column];
		for(int i=0;i<rowNum;i++)
			for(int j=0;j<column;j++)
				lenth[i][j]=1;
		int count;
		int max=0;
		for(int k=0;k<column*rowNum-1;k++){
			count=0;
			for(int i=0;i<rowNum;i++){
				for(int j=0;j<column;j++){
					if(i-1>=0&&matrix[i-1][j]<matrix[i][j]){
						lenth[i][j]=lenth[i-1][j]+1;
						count++;
						max=max<lenth[i][j]?lenth[i][j]:max;
					}
					if(j-1>=0&&matrix[i][j-1]<matrix[i][j]){
						lenth[i][j]=lenth[i][j-1]+1;
						count++;
						max=max<lenth[i][j]?lenth[i][j]:max;
					}
					
					if(i+1<rowNum&&matrix[i+1][j]<matrix[i][j]){
						lenth[i][j]=lenth[i+1][j]+1;
						count++;
						max=max<lenth[i][j]?lenth[i][j]:max;
					}
					if(j+1<column&&matrix[i][j+1]<matrix[i][j]){
						lenth[i][j]=lenth[i][j+1]+1;
						count++;
						max=max<lenth[i][j]?lenth[i][j]:max;
					}
				}
			}
			if(count==0)
				break;
		}
		return max;
	}
}