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：
package main
import "fmt"
func solveSudoku(board [][]byte) {
rows := make([]map[byte]bool, 9) // 行集合
cols := make([]map[byte]bool, 9) // 列集合
sudo := make([]map[byte]bool, 9) // 3x3宫格集合
empty := make([][2]int, 0) // 空格子坐标
// 初始化行、列、3x3宫格集合
for i := 0; i < 9; i++ {
rows[i] = make(map[byte]bool)
cols[i] = make(map[byte]bool)
sudo[i] = make(map[byte]bool)
for j := 1; j <= 9; j++ {
rows[i][byte(j+'0')] = true
cols[i][byte(j+'0')] = true
sudo[i][byte(j+'0')] = true
}
}
// 预处理，将空格子坐标和行、列、3x3宫格集合更新
for i := 0; i < 9; i++ {
for j := 0; j < 9; j++ {
if board[i][j] != '.' {
val := board[i][j]
delete(rows[i], val)
delete(cols[j], val)
delete(sudo[(i/3)*3+j/3], val)
} else {
empty = append(empty, [2]int{i, j})
}
}
}
// 回溯函数
var backtrack func(int) bool
backtrack = func(iter int) bool {
// 所有空格子坐标均已填充
if iter == len(empty) {
return true
}
// 获取空格子坐标
i, j := empty[iter][0], empty[iter][1]
// 获取该格子所在3x3宫格索引
k := (i/3)*3 + j/3
// 该格子可选数值集合
choices := make(map[byte]bool)
for num := range rows[i] {
if cols[j][num] && sudo[k][num] {
choices[num] = true
}
}
for num := range choices {
// 尝试填充该格子
board[i][j] = num
// 更新行、列、3x3宫格集合
delete(rows[i], num)
delete(cols[j], num)
delete(sudo[k], num)
// 递归填充下一个空格子
if backtrack(iter + 1) {
return true
}
// 回溯，还原现场
rows[i][num] = true
cols[j][num] = true
sudo[k][num] = true
}
// 所有数值均不可选，回溯到上一层
board[i][j] = '.'
return false
}
backtrack(0)
}
func main() {
board := [][]byte{
{'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'}}
solveSudoku(board)
for _, row := range board {
for _, col := range row {
fmt.Print(col-48, " ")
}
fmt.Println()
}
answer := [][]byte{
{'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'}}
// 判断与答案是否一致
equal := true
for i, row := range board {
for j, col := range row {
if col != answer[i][j] {
equal = false
break
}
}
if !equal {
break
}
}
fmt.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
代码2：
package main
import "fmt"
type position struct {
x int
y int
}
func solveSudoku(board [][]byte) {
pos, find := []position{}, false
for i := 0; i < len(board); i++ {
for j := 0; j < len(board[0]); j++ {
if board[i][j] == '.' {
pos = append(pos, position{x: i, y: j})
}
}
}
putSudoku(&board, pos, 0, &find)
}
func putSudoku(board *[][]byte, pos []position, index int, succ *bool) {
if *succ == true {
return
}
if index == len(pos) {
*succ = true
return
}
for i := 1; i < 10; i++ {
if checkSudoku(board, pos[index], i) && !*succ {
(*board)[pos[index].x][pos[index].y] = byte(i) + '0'
putSudoku(board, pos, index+1, succ)
if *succ == true {
return
}
(*board)[pos[index].x][pos[index].y] = '.'
}
}
}
func checkSudoku(board *[][]byte, pos position, val int) bool {
// 判断行是否有重复数字
for i := 0; i < len((*board)[0]); i++ {
if (*board)[pos.x][i] != '.' && int((*board)[pos.x][i]-'0') == val {
return false
}
}
// 判断列是否有重复数字
for i := 0; i < len((*board)); i++ {
if (*board)[i][pos.y] != '.' && int((*board)[i][pos.y]-'0') == val {
return false
}
}
// 判断九宫格是否有重复数字
posx, posy := pos.x-pos.x%3, pos.y-pos.y%3
for i := posx; i < posx+3; i++ {
for j := posy; j < posy+3; j++ {
if (*board)[i][j] != '.' && int((*board)[i][j]-'0') == val {
return false
}
}
}
return true
}
func main() {
board := [][]byte{
{'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'}}
solveSudoku(board)
for _, row := range board {
for _, col := range row {
fmt.Print(col-48, " ")
}
fmt.Println()
}
answer := [][]byte{
{'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'}}
// 判断与答案是否一致
equal := true
for i, row := range board {
for j, col := range row {
if col != answer[i][j] {
equal = false
break
}
}
if !equal {
break
}
}
fmt.Println(equal)
}
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：
package main
import (
"fmt"
"strconv"
"strings"
)
func countAndSay(n int) string {
if n == 1 {
return "1"
}
prev := countAndSay(n - 1)
var res strings.Builder
for i, j := 0, 0; j <= len(prev); j++ {
if j == len(prev) || prev[j] != prev[i] {
res.WriteString(strconv.Itoa(j - i))
res.WriteByte(prev[i])
i = j
}
}
return res.String()
}
func main() {
for i := 1; i < 6; i++ {
fmt.Println(countAndSay(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：
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：
package main
import (
"fmt"
"sort"
)
func combinationSum(candidates []int, target int) [][]int {
if len(candidates) == 0 {
return [][]int{}
}
c, res := []int{}, [][]int{}
sort.Ints(candidates)
findcombinationSum(candidates, target, 0, c, &res)
return res
}
func findcombinationSum(nums []int, target, index int, c []int, res *[][]int) {
if target <= 0 {
if target == 0 {
b := make([]int, len(c))
copy(b, c)
*res = append(*res, b)
}
return
}
for i := index; i < len(nums); i++ {
if nums[i] > target {
break
}
c = append(c, nums[i])
findcombinationSum(nums, target-nums[i], i, c, res)
c = c[:len(c)-1]
}
}
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))
}
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