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.

题目大意 #

给定一个整数矩阵，找出最长递增路径的长度。对于每个单元格，你可以往上，下，左，右四个方向移动。你不能在对角线方向上移动或移动到边界外（即不允许环绕）。

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"
)

var dir = [][]int{
	{-1, 0},
	{0, 1},
	{1, 0},
	{0, -1},
}

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 max(a int, b int) int {
	if a > b {
		return a
	}
	return b
}

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
}

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