leetcode: Longest Increasing Path in a Matrix

找出一个矩阵中最长的递增序列，求出最长长度是多少

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:

nums = [

[9,9,4],

[6,6,8],

[2,1,1]

]

Return 4

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

Example 2:

nums = [

[3,4,5],

[3,2,6],

[2,2,1]

]

Return 4

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

直接深度优先搜索，入门题目

public class Solution {

static int[][] matrix;
static int[][] visited;
static int length, width;
static int[][] step = { { 0, -1 }, { 0, 1 }, { -1, 0 }, { 1, 0 } };

public int longestIncreasingPath(int[][] matrix) {
if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
return 0;
}
else if(matrix.length == 1 && matrix[0].length == 1){
return 1;
}
length = matrix.length;
width = matrix[0].length;
Solution.matrix = matrix;
Solution.visited = new int[length][width];
int maxLength = 1;
for (int i = 0; i < visited.length; ++i) {
for (int j = 0; j < visited[i].length; ++j) {
dfs(i, j, 1);
if (maxLength < visited[i][j]) {
maxLength = visited[i][j];
}
}
}
return maxLength;
}

public static void dfs(int x, int y, int curLength) {
if (visited[x][y] != 0) {
return;
}
for (int t = 0; t < step.length; ++t) {
int i = x + step[t][0], j = y + step[t][1];
if (!check(i, j)) {
continue;
}
if (matrix[x][y] > matrix[i][j]) {
if (visited[i][j] == 0) {
dfs(i, j, 1);
}
curLength = Math.max(curLength, visited[i][j] + 1);
}
}
visited[x][y] = curLength;
}

public static boolean check(int x, int y) {
return x >= 0 && x < length && y >= 0 && y < width;
}
}