README.md

329. Longest Increasing Path in a Matrix

Given an m x n integers matrix, return the length of the longest increasing path in matrix.

From each cell, you can either move in four directions: left, right, up, or down. You may not move diagonally or move outside the boundary (i.e., wrap-around is not allowed).

Example 1:

Input: matrix = [[9,9,4],[6,6,8],[2,1,1]]
Output: 4
Explanation: The longest increasing path is [1, 2, 6, 9].

Example 2:

Input: matrix = [[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.

Example 3:

Input: matrix = [[1]]
Output: 1

Constraints:

• m == matrix.length
• n == matrix[i].length
• 1 <= m, n <= 200
• 0 <= matrix[i][j] <= 231 - 1

Solutions (Rust)

1. Solution

impl Solution {
    pub fn longest_increasing_path(matrix: Vec<Vec<i32>>) -> i32 {
        let m = matrix.len();
        let n = matrix[0].len();
        let mut cells = vec![];
        let mut dp = vec![vec![1; n]; m];
        let mut ret = 1;

        for r in 0..m {
            for c in 0..n {
                cells.push((r, c));
            }
        }

        cells.sort_unstable_by_key(|&(r, c)| matrix[r][c]);

        for (r, c) in cells {
            let mut max_path = 0;

            if r > 0 && matrix[r - 1][c] < matrix[r][c] {
                max_path = max_path.max(dp[r - 1][c]);
            }
            if r < m - 1 && matrix[r + 1][c] < matrix[r][c] {
                max_path = max_path.max(dp[r + 1][c]);
            }
            if c > 0 && matrix[r][c - 1] < matrix[r][c] {
                max_path = max_path.max(dp[r][c - 1]);
            }
            if c < n - 1 && matrix[r][c + 1] < matrix[r][c] {
                max_path = max_path.max(dp[r][c + 1]);
            }

            dp[r][c] = max_path + 1;
            ret = ret.max(dp[r][c]);
        }

        ret
    }
}