目录
37. 解数独 Sudoku Solver 🌟🌟🌟
38. 外观数列 Count and Say 🌟🌟
39. 组合总和 Combination Sum 🌟🌟
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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 == 9board[i].length == 9board[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 <= 301 <= candidates[i] <= 200candidate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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