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.
Solution:
func longestIncreasingPath(matrix [][]int) int {
row := len(matrix)
if row == 0 {
return 0
}
col := len(matrix[0])
if col == 0 {
return 0
}
dirs := [4][2]int{
[2]int{1, 0},
[2]int{-1, 0},
[2]int{0, 1},
[2]int{0, -1},
}
cache := make([][]int, row)
for i := range cache {
cache[i] = make([]int, col)
}
var dfs func(i, j int) int
dfs = func(i, j int) int {
if cache[i][j] != 0 {
return cache[i][j]
}
for _, d := range dirs {
tmp_i, tmp_j := i + d[0], j + d[1]
if tmp_i >= 0 && tmp_i < row && tmp_j >= 0 && tmp_j < col && matrix[i][j] < matrix[tmp_i][tmp_j] {
cache[i][j] = max(cache[i][j], dfs(tmp_i, tmp_j))
}
}
cache[i][j]++
return cache[i][j]
}
var ans int
for i := 0; i < row; i++ {
for j := 0; j < col; j++ {
ans = max(ans, dfs(i, j))
}
}
return ans
}
func max(a, b int) int {
if a > b {
return a
}
return b
}