37. 解数独 Sudoku Solver
编写一个程序,通过填充空格来解决数独问题。
数独的解法需 遵循如下规则:
- 数字
1-9
在每一行只能出现一次。 - 数字
1-9
在每一列只能出现一次。 - 数字
1-9
在每一个以粗实线分隔的3x3
宫内只能出现一次。(请参考示例图)
数独部分空格内已填入了数字,空白格用 '.'
表示。
示例 1:
输入:board = [
["5","3",".",".","7",".",".",".","."],
["6",".",".","1","9","5",".",".","."],
[".","9","8",".",".",".",".","6","."],
["8",".",".",".","6",".",".",".","3"],
["4",".",".","8",".","3",".",".","1"],
["7",".",".",".","2",".",".",".","6"],
[".","6",".",".",".",".","2","8","."],
[".",".",".","4","1","9",".",".","5"],
[".",".",".",".","8",".",".","7","9"]]
输出:
[["5","3","4","6","7","8","9","1","2"],
["6","7","2","1","9","5","3","4","8"],
["1","9","8","3","4","2","5","6","7"],
["8","5","9","7","6","1","4","2","3"],
["4","2","6","8","5","3","7","9","1"],
["7","1","3","9","2","4","8","5","6"],
["9","6","1","5","3","7","2","8","4"],
["2","8","7","4","1","9","6","3","5"],
["3","4","5","2","8","6","1","7","9"]]
解释:输入的数独如上图所示,唯一有效的解决方案如下所示:
提示:
board.length == 9
board[i].length == 9
board[i][j]
是一位数字或者'.'
- 题目数据 保证 输入数独仅有一个解
代码:
fn solve_sudoku(board: &mut Vec<Vec<char>>) { let mut pos: Vec<[usize; 2]> = Vec::new(); let mut find = false; for i in 0..board.len() { for j in 0..board[0].len() { if board[i][j] == '.' { pos.push([i, j]); } } } put_sudoku(board, &pos, 0, &mut find); } fn put_sudoku( board: &mut Vec<Vec<char>>, pos: &Vec<[usize; 2]>, index: usize, succ: &mut bool, ) { if *succ { return; } if index == pos.len() { *succ = true; return; } for i in 1..=9 { if check_sudoku(board, pos[index], i) && !*succ { board[pos[index][0]][pos[index][1]] = (i as u8 + b'0') as char; put_sudoku(board, pos, index + 1, succ); if *succ { return; } board[pos[index][0]][pos[index][1]] = '.'; } } } fn check_sudoku(board: &Vec<Vec<char>>, pos: [usize; 2], val: usize) -> bool { // 判断行是否有重复数字 for i in 0..board[0].len() { if board[pos[0]][i] != '.' && (board[pos[0]][i] as u8 - b'0') as usize == val { return false; } } // 判断列是否有重复数字 for i in 0..board.len() { if board[i][pos[1]] != '.' && (board[i][pos[1]] as u8 - b'0') as usize == val { return false; } } // 判断九宫格是否有重复数字 let posx = pos[0] - pos[0] % 3; let posy = pos[1] - pos[1] % 3; for i in posx..posx + 3 { for j in posy..posy + 3 { if board[i][j] != '.' && (board[i][j] as u8 - b'0') as usize == val { return false; } } } true } fn main() { let mut board: Vec<Vec<char>> = vec![ vec!['5', '3', '.', '.', '7', '.', '.', '.', '.'], vec!['6', '.', '.', '1', '9', '5', '.', '.', '.'], vec!['.', '9', '8', '.', '.', '.', '.', '6', '.'], vec!['8', '.', '.', '.', '6', '.', '.', '.', '3'], vec!['4', '.', '.', '8', '.', '3', '.', '.', '1'], vec!['7', '.', '.', '.', '2', '.', '.', '.', '6'], vec!['.', '6', '.', '.', '.', '.', '2', '8', '.'], vec!['.', '.', '.', '4', '1', '9', '.', '.', '5'], vec!['.', '.', '.', '.', '8', '.', '.', '7', '9'], ]; solve_sudoku(&mut board); for row in &board { for col in row { print!("{} ", (*col as u8 - b'0') as usize); } println!(); } let answer: Vec<Vec<char>> = vec![ vec!['5', '3', '4', '6', '7', '8', '9', '1', '2'], vec!['6', '7', '2', '1', '9', '5', '3', '4', '8'], vec!['1', '9', '8', '3', '4', '2', '5', '6', '7'], vec!['8', '5', '9', '7', '6', '1', '4', '2', '3'], vec!['4', '2', '6', '8', '5', '3', '7', '9', '1'], vec!['7', '1', '3', '9', '2', '4', '8', '5', '6'], vec!['9', '6', '1', '5', '3', '7', '2', '8', '4'], vec!['2', '8', '7', '4', '1', '9', '6', '3', '5'], vec!['3', '4', '5', '2', '8', '6', '1', '7', '9'], ]; // 判断与答案是否一致 let mut equal = true; for i in 0..board.len() { for j in 0..board[0].len() { if board[i][j] != answer[i][j] { equal = false; break; } } if !equal { break; } } println!("{}", equal); }
输出:
5 3 4 6 7 8 9 1 2
6 7 2 1 9 5 3 4 8
1 9 8 3 4 2 5 6 7
8 5 9 7 6 1 4 2 3
4 2 6 8 5 3 7 9 1
7 1 3 9 2 4 8 5 6
9 6 1 5 3 7 2 8 4
2 8 7 4 1 9 6 3 5
3 4 5 2 8 6 1 7 9
true
38. 外观数列 Count and Say
给定一个正整数 n
,输出外观数列的第 n
项。
「外观数列」是一个整数序列,从数字 1 开始,序列中的每一项都是对前一项的描述。
你可以将其视作是由递归公式定义的数字字符串序列:
countAndSay(1) = "1"
countAndSay(n)
是对countAndSay(n-1)
的描述,然后转换成另一个数字字符串。
前五项如下:
1. 1
2. 11
3. 21
4. 1211
5. 111221
第一项是数字 1 描述前一项,这个数是 1 即 “ 一 个 1 ”,记作 "11"
描述前一项,这个数是 11 即 “ 二 个 1 ” ,记作 "21"
描述前一项,这个数是 21 即 “ 一 个 2 + 一 个 1 ” ,记作 "1211"
描述前一项,这个数是 1211 即 “ 一 个 1 + 一 个 2 + 二 个 1 ” ,记作 "111221"
要 描述 一个数字字符串,首先要将字符串分割为 最小 数量的组,每个组都由连续的最多 相同字符 组成。然后对于每个组,先描述字符的数量,然后描述字符,形成一个描述组。要将描述转换为数字字符串,先将每组中的字符数量用数字替换,再将所有描述组连接起来。
例如,数字字符串 "3322251"
的描述如下图:
示例 1:
输入:n = 1
输出:"1"
解释:这是一个基本样例。
示例 2:
输入:n = 4
输出:"1211"
解释:
countAndSay(1) = "1"
countAndSay(2) = 读 "1" = 一 个 1 = "11"
countAndSay(3) = 读 "11" = 二 个 1 = "21"
countAndSay(4) = 读 "21" = 一 个 2 + 一 个 1 = "12" + "11" = "1211"
提示:
1 <= n <= 30
代码:
fn count_and_say(n: i32) -> String { if n == 1 { return String::from("1"); } let prev = count_and_say(n - 1); let mut res = String::new(); let mut i = 0; let mut j = 0; while j <= prev.len() { if j == prev.len() || prev.chars().nth(j) != prev.chars().nth(i) { res += &(j - i).to_string(); res += &prev.chars().nth(i).unwrap().to_string(); i = j; } j += 1; } res } fn main() { for i in 1..=5 { println!("{}", count_and_say(i)); } }
输出:
1
11
21
1211
111221
39. 组合总和 Combination Sum
给你一个 无重复元素 的整数数组 candidates
和一个目标整数 target
,找出 candidates
中可以使数字和为目标数 target
的 所有 不同组合 ,并以列表形式返回。你可以按 任意顺序 返回这些组合。
candidates
中的 同一个 数字可以 无限制重复被选取 。如果至少一个数字的被选数量不同,则两种组合是不同的。
对于给定的输入,保证和为 target
的不同组合数少于 150
个。
示例 1:
输入:candidates = [2,3,6,7], target = 7
输出:[[2,2,3],[7]]
解释:
2 和 3 可以形成一组候选,2 + 2 + 3 = 7 。注意 2 可以使用多次。
7 也是一个候选, 7 = 7 。仅有这两种组合。
示例 2:
输入: candidates = [2,3,5], target = 8
输出: [[2,2,2,2],[2,3,3],[3,5]]
示例 3:
输入: candidates = [2], target = 1
输出: []
提示:
1 <= candidates.length <= 30
1 <= candidates[i] <= 200
candidate
中的每个元素都 互不相同1 <= target <= 500
代码1: 回溯法
package main import "fmt" func combinationSum(candidates []int, target int) [][]int { var res [][]int var backtrack func([]int, int, int) backtrack = func(path []int, sum int, start int) { if sum >= target { if sum == target { res = append(res, append([]int{}, path...)) return } return } for i := start; i < len(candidates); i++ { path = append(path, candidates[i]) backtrack(path, sum+candidates[i], i) path = path[:len(path)-1] } } backtrack([]int{}, 0, 0) return res } func main() { candidates := []int{2, 3, 6, 7} fmt.Println(combinationSum(candidates, 7)) candidates = []int{2, 3, 5} fmt.Println(combinationSum(candidates, 8)) candidates = []int{2} fmt.Println(combinationSum(candidates, 1)) }
输出:
[[2, 2, 3], [7]]
[[2, 2, 2, 2], [2, 3, 3], [3, 5]]
[]
代码2: 回溯法
fn combination_sum(candidates: Vec<i32>, target: i32) -> Vec<Vec<i32>> { if candidates.is_empty() { return Vec::new(); } let mut res = Vec::new(); let mut c = Vec::new(); let mut nums = candidates.clone(); nums.sort(); backtrack(&nums, target, 0, &mut c, &mut res); res } fn backtrack(nums: &Vec<i32>, target: i32, index: usize, c: &mut Vec<i32>, res: &mut Vec<Vec<i32>>) { if target <= 0 { if target == 0 { res.push(c.clone()); } return; } for i in index..nums.len() { if nums[i] > target { break; } c.push(nums[i]); backtrack(nums, target - nums[i], i, c, res); c.pop(); } } fn main() { let candidates = vec![2, 3, 6, 7]; println!("{:?}", combination_sum(candidates, 7)); let candidates = vec![2, 3, 5]; println!("{:?}", combination_sum(candidates, 8)); let candidates = vec![2]; println!("{:?}", combination_sum(candidates, 1)); }
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