LeetCode 329 Longest Increasing Path in a Matrix (记忆化搜索)

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.

Credits:

Special thanks to
@dietpepsi for adding this problem and creating all test cases.

题目链接：https://leetcode.com/problems/longest-increasing-path-in-a-matrix/

题目分析：设dp[i][j]为到(i, j)这个位置时最长上升路径长度，击败了50%。。。

public class Solution {

int [][]dp;
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};

int DFS (int x, int y, int n, int m, int[][] matrix) {
if (dp[x][y] != 0) {
return dp[x][y];
}
dp[x][y] = 1;
for (int i = 0; i < 4; i ++) {
int xx = x + dx[i];
int yy = y + dy[i];
if (xx >= 0 && xx < n && yy >= 0 && yy < m && matrix[xx][yy] < matrix[x][y]) {
dp[x][y] = Math.max(dp[x][y], DFS(xx, yy, n, m, matrix) + 1);
}
}
return dp[x][y];
}

public int longestIncreasingPath(int[][] matrix) {
int n = matrix.length;
if (n == 0) {
return 0;
}
int m = matrix[0].length;
dp = new int[n + 1][m + 1];
for (int i = 0; i < n; i ++) {
Arrays.fill(dp[i], 0);
}
int ans = 1;
for (int i = 0; i < n; i ++) {
for (int j = 0; j < m; j ++) {
ans = Math.max(ans, DFS(i, j, n, m, matrix));
}
}
return ans;
}
}