[代码随想录]Day26-回溯算法part06

发布时间 2023-08-24 10:31:29作者: WtcSky

题目:332. 重新安排行程

思路:

其实这里已经是图的部分了,回溯应该也可以。Hierholzer算法解决欧拉问题

代码:

func findItinerary(tickets [][]string) []string {
    var (
        m  = map[string][]string{}
        res []string
    )
    
    for _, ticket := range tickets {
        src, dst := ticket[0], ticket[1]
        m[src] = append(m[src], dst)
    }
    for key := range m {
        sort.Strings(m[key])
    }

    var dfs func(curr string)
    dfs = func(curr string) {
        for {
            if v, ok := m[curr]; !ok || len(v) == 0 {
                break
            }
            tmp := m[curr][0]
            m[curr] = m[curr][1:]
            dfs(tmp)
        }
        res = append(res, curr)
    }

    dfs("JFK")
    for i := 0; i < len(res)/2; i++ {
        res[i], res[len(res) - 1 - i] = res[len(res) - 1 - i], res[i]
    }
    return res
}

参考:

代码随想录

题目:51. N 皇后

思路:

哈哈哈,先这样吧

代码:

var solutions [][]string

func solveNQueens(n int) [][]string {
    solutions = [][]string{}
    queens := make([]int, n)
    for i := 0; i < n; i++ {
        queens[i] = -1
    }
    columns := map[int]bool{}
    diagonals1, diagonals2 := map[int]bool{}, map[int]bool{}
    backtrack(queens, n, 0, columns, diagonals1, diagonals2)
    return solutions
}

func backtrack(queens []int, n, row int, columns, diagonals1, diagonals2 map[int]bool) {
    if row == n {
        board := generateBoard(queens, n)
        solutions = append(solutions, board)
        return
    }
    for i := 0; i < n; i++ {
        if columns[i] {
            continue
        }
        diagonal1 := row - i
        if diagonals1[diagonal1] {
            continue
        }
        diagonal2 := row + i
        if diagonals2[diagonal2] {
            continue
        }
        queens[row] = i
        columns[i] = true
        diagonals1[diagonal1], diagonals2[diagonal2] = true, true
        backtrack(queens, n, row + 1, columns, diagonals1, diagonals2)
        queens[row] = -1
        delete(columns, i)
        delete(diagonals1, diagonal1)
        delete(diagonals2, diagonal2)
    }
}

func generateBoard(queens []int, n int) []string {
    board := []string{}
    for i := 0; i < n; i++ {
        row := make([]byte, n)
        for j := 0; j < n; j++ {
            row[j] = '.'
        }
        row[queens[i]] = 'Q'
        board = append(board, string(row))
    }
    return board
}

参考:

代码随想录

题目:37. 解数独

思路:

哈哈哈,先这样吧

代码:

func solveSudoku(board [][]byte) {
	var backtracking func(board [][]byte) bool
	backtracking = func(board [][]byte) bool {
		for i := 0; i < 9; i++ {
			for j := 0; j < 9; j++ {
				//判断此位置是否适合填数字
				if board[i][j] != '.' {
					continue
				}
				//尝试填1-9
				for k := '1'; k <= '9'; k++ {
					if isvalid(i, j, byte(k), board) == true { //如果满足要求就填
						board[i][j] = byte(k)
						if backtracking(board) == true {
							return true
						}
						board[i][j] = '.'
					}
				}
				return false
			}
		}
		return true
	}
	backtracking(board)
}

//判断填入数字是否满足要求
func isvalid(row, col int, k byte, board [][]byte) bool {
	for i := 0; i < 9; i++ { //行
		if board[row][i] == k {
			return false
		}
	}
	for i := 0; i < 9; i++ { //列
		if board[i][col] == k {
			return false
		}
	}
	//方格
	startrow := (row / 3) * 3
	startcol := (col / 3) * 3
	for i := startrow; i < startrow+3; i++ {
		for j := startcol; j < startcol+3; j++ {
			if board[i][j] == k {
				return false
			}
		}
	}
	return true
}

参考:

代码随想录