Golang每日一练(leetDay0013)

目录

37. 解数独 Sudoku Solver  🌟🌟🌟

38. 外观数列 Count and Say  🌟🌟

39. 组合总和 Combination Sum  🌟🌟

🌟 每日一练刷题专栏 🌟

Golang每日一练 专栏

Python每日一练 专栏

C/C++每日一练 专栏

Java每日一练 专栏

---

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：

1. 数字 1-9 在每一行只能出现一次。
2. 数字 1-9 在每一列只能出现一次。
3. 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：

package mainimport "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')] = truecols[i][byte(j+'0')] = truesudo[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) boolbacktrack = 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] = truecols[j][num] = truesudo[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 := truefor i, row := range board {for j, col := range row {if col != answer[i][j] {equal = falsebreak}}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 mainimport "fmt"type position struct {x inty int
}func solveSudoku(board [][]byte) {pos, find := []position{}, falsefor 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 = truereturn}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%3for 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 := truefor i, row := range board {for j, col := range row {if col != answer[i][j] {equal = falsebreak}}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 mainimport ("fmt""strconv""strings"
)func countAndSay(n int) string {if n == 1 {return "1"}prev := countAndSay(n - 1)var res strings.Builderfor 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 <= 30
• 1 <= candidates[i] <= 200
• candidate 中的每个元素都 互不相同
• 1 <= target <= 500

代码：

package mainimport "fmt"func combinationSum(candidates []int, target int) [][]int {var res [][]intvar 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 mainimport ("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))}

---

🌟 每日一练刷题专栏 🌟

✨ 持续，努力奋斗做强刷题搬运工！

👍 点赞，你的认可是我坚持的动力！ 

🌟 收藏，你的青睐是我努力的方向！ 

✎ 评论，你的意见是我进步的财富！  

	Golang每日一练 专栏
	Python每日一练 专栏
	C/C++每日一练 专栏
	Java每日一练 专栏