以下是 LeetCode 3459. 最长 V 形对角线段的长度 的 Rust 实现rustuse std::collections::HashMap;impl Solution {pub fn len_of_v_diagonal(grid: VecVeci32) - i32 {let n grid.len();if n 0 { return 0; }let m grid[0].len();// 四个对角线方向顺时针排列// 0: 右上 (-1, 1)// 1: 右下 (1, 1)// 2: 左下 (1, -1)// 3: 左上 (-1, -1)let dirs: [(isize, isize); 4] [(-1, 1), (1, 1), (1, -1), (-1, -1)];// 记忆化缓存: (i, j, turned, num, dir_idx) - 最长路径长度let mut memo: HashMap(usize, usize, bool, i32, usize), i32 HashMap::new();fn dfs(grid: VecVeci32,dirs: [(isize, isize); 4],memo: mut HashMap(usize, usize, bool, i32, usize), i32,i: usize,j: usize,turned: bool,num: i32,dir_idx: usize,n: usize,m: usize,) - i32 {// 越界检查if i n || j m {return 0;}// 值不匹配if grid[i][j] ! num {return 0;}// 查缓存let key (i, j, turned, num, dir_idx);if let Some(val) memo.get(key) {return val;}// 下一个期望值2和0交替let next_num if num 2 { 0 } else { 2 };let (dx, dy) dirs[dir_idx];// 选择1继续沿原方向走let ni (i as isize dx) as usize;let nj (j as isize dy) as usize;let mut res 1 dfs(grid, dirs, memo, ni, nj, turned, next_num, dir_idx, n, m);// 选择2如果没转过弯顺时针转90度if !turned {let next_dir (dir_idx 1) % 4;let (ndx, ndy) dirs[next_dir];let nni (i as isize ndx) as usize;let nnj (j as isize ndy) as usize;res res.max(1 dfs(grid, dirs, memo, nni, nnj, true, next_num, next_dir, n, m));}memo.insert(key, res);res}let mut ans 0;for i in 0..n {for j in 0..m {if grid[i][j] 1 {// 从每个1出发尝试4个方向for d in 0..4 {let (dx, dy) dirs[d];let ni (i as isize dx) as usize;let nj (j as isize dy) as usize;ans ans.max(1 dfs(grid, dirs, mut memo, ni, nj, false, 2, d, n, m));}}}}ans}}核心要点- 越界处理利用 Rust 中 usize 的无符号特性当 i dx 0 时as usize 会自然溢出为很大的数从而被 i n 捕获无需额外判断负数- 记忆化使用 HashMap 缓存 (i, j, turned, num, dir_idx) 状态避免重复计算- 方向数组4 个对角线方向按顺时针排列转弯即 (dir_idx 1) % 4- 序列规则1 → 2 → 0 → 2 → 0 → ...通过 next_num if num 2 { 0 } else { 2 } 实现