(1)参考
创建map存储src,[]dest映射关系,并对[]dest排序
每次取map中第一个dest访问,将其作为新的src,每访问一条src->dest,删除该记录。
如果访问的src没有dest了,将当前节点加入结果集,并沿栈返回。
结果是沿栈返回的,故需要逆序输出
func findItinerary(tickets [][]string) []string {
res := []string{}
m := make(map[string][]string, 0)
for _, ticket := range tickets {
src, dest := ticket[0], ticket[1]
m[src] = append(m[src], dest)
}
for k:= range m {
sort.Strings(m[k])
}
var help func(srcTicket string)
help = func(srcTicket string) {
for {
if v, ok := m[srcTicket]; !ok || len(v) == 0 {
break
}
tmp := m[srcTicket][0]
m[srcTicket] = m[srcTicket][1:]
help(tmp)
}
res = append(res, srcTicket)
}
help("JFK")
for i, j := 0, len(res)-1; i < j; i, j = i+1, j-1 {
res[i], res[j] = res[j], res[i]
}
return res
}
(2)回溯:超时了
排列问题。先对tickets排序,used记录当前车票是否被使用
若车票使用完并找到路径,返回,否则回溯查找路径
func findItinerary(tickets [][]string) []string {
sort.Slice(tickets, func(i, j int) bool {
return tickets[i][1] < tickets[j][1]
})
path := []string{"JFK"}
used := make([]bool, len(tickets))
var help func(srcTicket string, ticket [][]string) bool
help = func(srcTicket string, ticket [][]string) bool {
if len(path) == len(tickets)+1 {
return true
}
for i, next := range tickets {
if next[0] == path[len(path)-1] && !used[i] {
path = append(path, next[1])
used[i] = true
if help(next[1], tickets) {
return true
}
path = path[:len(path)-1]
used[i] = false
}
}
return false
}
help("JFK", tickets)
return path
}
回溯
row控制递归深度,for循环控制列,进而确定当前位置
判断当前值是否有效:因为每行只选取一个位置,故只需判断列和正斜、反斜方向是否有皇后
func solveNQueens(n int) [][]string {
res := [][]string{}
board := make([][]string, n)
for i := 0; i < n; i++ {
board[i] = make([]string, n)
}
for i := 0; i < n; i++ {
for j := 0; j < n; j++ {
board[i][j] = "."
}
}
var help func(row int, board [][]string) bool
help = func(row int, board [][]string) bool {
if row == n {
temp := make([]string, n)
for i, rowStr := range board {
temp[i] = strings.Join(rowStr, "")
}
res = append(res, temp)
}
for i := 0; i < n; i++ {
if isValid(n, row, i, board) {
board[row][i] = "Q"
if help(row+1, board) {
return true
}
board[row][i] = "."
}
}
return false
}
help(0, board)
return res
}
func isValid(n, row, col int, board [][]string) bool {
for i := 0; i < row; i++ {
if board[i][col] == "Q" {
return false
}
}
for i, j := row-1, col-1; i >= 0 && j >= 0; i, j = i-1, j-1 {
if board[i][j] == "Q" {
return false
}
}
for i, j := row-1, col+1; i >= 0 && j < n; i, j = i-1, j+1 {
if board[i][j] == "Q" {
return false
}
}
return true
}
回溯
当前位置有效性判断:行、列、九宫格数字不重复
如果当前位置能放数字,且有效,递归,否则回溯
func solveSudoku(board [][]byte) {
var help func(board [][]byte) bool
help = func(board [][]byte) bool {
for i := 0; i < 9; i++ {
for j := 0; j < 9; j++ {
if board[i][j] == '.' {
for k := '1'; k <= '9'; k++ {
if isValid(i, j, byte(k), board) {
board[i][j] = byte(k)
if help(board) {
return true
}
board[i][j] = '.'
}
}
return false
}
}
}
return true
}
help(board)
}
func isValid(row, col int, val byte, board [][]byte) bool {
for i := 0; i < 9; i++ {
if board[row][i] == val {
return false
}
}
for i := 0; i < 9; i++ {
if board[i][col] == val {
return false
}
}
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] == val {
return false
}
}
}
return true
}
332.重新安排行程
51.N皇后
37.解数独
回溯总结篇