0329. Longest Increasing Path in a Matrix

# 329. Longest Increasing Path in a Matrix#

## 题目 #

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.
``````

## 解题思路 #

• 给出一个矩阵，要求在这个矩阵中找到一个最长递增的路径。路径有上下左右 4 个方向。
• 这一题解题思路很明显，用 DFS 即可。在提交完第一版以后会发现 TLE，因为题目给出了一个非常大的矩阵，搜索次数太多。所以需要用到记忆化，把曾经搜索过的最大长度缓存起来，增加了记忆化以后再次提交 AC。

## 代码 #

``````
package leetcode

import (
"math"
)

func longestIncreasingPath(matrix [][]int) int {
cache, res := make([][]int, len(matrix)), 0
for i := 0; i < len(cache); i++ {
cache[i] = make([]int, len(matrix[0]))
}
for i, v := range matrix {
for j := range v {
searchPath(matrix, cache, math.MinInt64, i, j)
res = max(res, cache[i][j])
}
}
return res
}

func isInIntBoard(board [][]int, x, y int) bool {
return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
}

func searchPath(board, cache [][]int, lastNum, x, y int) int {
if board[x][y] <= lastNum {
return 0
}
if cache[x][y] > 0 {
return cache[x][y]
}
count := 1
for i := 0; i < 4; i++ {
nx := x + dir[i][0]
ny := y + dir[i][1]
if isInIntBoard(board, nx, ny) {
count = max(count, searchPath(board, cache, board[x][y], nx, ny)+1)
}
}
cache[x][y] = count
return count
}

``````

Sep 6, 2020