Longest Increasing Path in a Matrix
Description
Given an m x n
integers matrix
, return the length of the longest increasing path in matrix
.
From each cell, you can either move in four directions: left, right, up, or down. You may not move diagonally or move outside the boundary (i.e., wrap-around is not allowed).
Example 1:
![](https://assets.leetcode.com/uploads/2021/01/05/grid1.jpg)
Input: matrix = [[9,9,4],[6,6,8],[2,1,1]]
Output: 4
Explanation: The longest increasing path is [1, 2, 6, 9]
.
Example 2:
![](https://assets.leetcode.com/uploads/2021/01/27/tmp-grid.jpg)
Input: matrix = [[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.
Example 3:
Input: matrix = [[1]] Output: 1
Constraints:
m == matrix.length
n == matrix[i].length
1 <= m, n <= 200
0 <= matrix[i][j] <= 231 - 1
Solution(javascript)
/**
* @param {number[][]} matrix
* @return {number}
*/
var longestIncreasingPath = function(matrix) {
let ylen = matrix.length, xlen = matrix[0].length, ans = 0,
memo = Array.from({length: ylen}, el => new Uint16Array(xlen))
const dfs = (y, x) => {
if (memo[y][x]) return memo[y][x]
let val = matrix[y][x]
memo[y][x] = 1 + Math.max(
y < ylen - 1 && matrix[y+1][x] < val ? dfs(y+1,x) : 0,
y > 0 && matrix[y-1][x] < val ? dfs(y-1,x) : 0,
x < xlen - 1 && matrix[y][x+1] < val ? dfs(y,x+1) : 0,
x > 0 && matrix[y][x-1] < val ? dfs(y,x-1) : 0)
return memo[y][x]
}
for (let i = 0; i < ylen; i++)
for (let j = 0; j < xlen; j++)
ans = Math.max(ans, dfs(i, j))
return ans
};