[Leetcode] [JavaScript] Longest increasing Path in a Matrix

Longest increasing Path in a Matrix

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 are not allowed).

Example 1:

Nums = [  [9, 9,4],  [6, 6,8],  [2,1, 1]]

Return4
The longest increasing path is [1, 2, 6, 9] .

Example 2:

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

Return4
The longest increasing path is [3, 4, 5, 6] . Moving diagonally is not allowed.

https://leetcode.com/problems/longest-increasing-path-in-a-matrix/

Look for the longest incremental path and return the length.

DFS recursive traversal finds the longest path, opening a hash table to record the nodes that have been visited, improving efficiency.

1 /**2 * @param {number[][]} matrix3 * @return {number}4  */5 varLongestincreasingpath =function(matrix) {6     varDirection = [{x:-1, y:0}, {x:1, y:0}, {x:0, Y:-1}, {x:0, y:1}];7     varDictionary = {}, max = 0;8      for(vari = 0; i < matrix.length; i++)9          for(varj = 0; J < Matrix[i].length; J + +)TenMax =Math.max (max, DFS (i, j)); One     returnMax; A      -     functiondfs (x, y) { -         varCurr = dictionary[x + ' # ' +y]; the         if(Curr)returnCurr; -Curr =Matrix[x][y]; -          -         varD, max = 0, isend =true, TMP; +          for(vari = 0; i < direction.length; i++){ -D =Direction[i]; +             if(Matrix[x +D.x]) { ATMP = matrix[x + d.x][y +D.y]; at                 if(TMP && tmp >Curr) { -max = Math.max (max, DFS (x + d.x, y +d.y)); -Isend =false; -                 } -             } -         } in          -max = Isend? 1:max + 1; toDictionary[x + ' # ' + y] =Max; +         returnMax; -     } the}

[Leetcode] [JavaScript] Longest increasing Path in a Matrix