Leetcode 38: how to reduce 3 kinds of situation to one solution?
$begingroup$
38. Count and Say
The count-and-say sequence is the sequence of integers with the first five terms as following:
1. 1
2. 11
3. 21
4. 1211
5. 111221
1is read off as"one 1"or11.
11is read off as"two 1s"or21.
21is read off as"one 2, then one 1"or1211.
Given an integer n where 1 ≤ n ≤ 30, generate the nth term of the count-and-say sequence.
Note: Each term of the sequence of integers will be represented as a string.
With iteration:
How to reduce 3 kinds of situation to one solution?
class Solution {
func countAndSay(_ n: Int) -> String {
var ans = ["1"]
var temp: [String], j: Int
for _ in 1..<n{
// from the beginning 1 to what we what
j = 1
temp =
for i in 1..<ans.count{
if ans[i - 1] == ans[i]{
j += 1
// Situation one: the current is equal to the previous ( util the last )
}
else{
temp.append(String(j))
temp.append(ans[i-1])
// Situation two: the current is not equal to the previous ( util the last )
j = 1
}
}
// Situation three: the last value
temp.append(String(j))
temp.append(ans.last!)
ans = temp
print(ans.reduce("", +))
}
return ans.reduce("", +)
}
}
With Recursion, which is a little leaner.
class Solution {
func countAndSay(_ n: Int) -> String {
if n == 1 {
return "1"
}
let str = countAndSay(n-1)
let arr = Array(str).map{Int(String($0))}
var last = arr[0]!
var temp : [(count: Int, val: Int)] = [(1, last)]
for num in arr[1...]{
if last == num{
var final = temp.last!
final.count += 1
temp.removeLast()
temp.append(final)
}
else{
temp += [(1, num!)]
}
last = num!
}
var result = ""
for key in temp{
result += "(key.0)(key.1)"
}
return result
}
}
How to go on doing some improvement?
programming-challenge swift
asked 11 mins ago
dengAprodengApro
176111
$endgroup$
add a comment |
$begingroup$
38. Count and Say
The count-and-say sequence is the sequence of integers with the first five terms as following:
1. 1
2. 11
3. 21
4. 1211
5. 111221
1is read off as"one 1"or11.
11is read off as"two 1s"or21.
21is read off as"one 2, then one 1"or1211.
Given an integer n where 1 ≤ n ≤ 30, generate the nth term of the count-and-say sequence.
Note: Each term of the sequence of integers will be represented as a string.
With iteration:
How to reduce 3 kinds of situation to one solution?
class Solution {
func countAndSay(_ n: Int) -> String {
var ans = ["1"]
var temp: [String], j: Int
for _ in 1..<n{
// from the beginning 1 to what we what
j = 1
temp =
for i in 1..<ans.count{
if ans[i - 1] == ans[i]{
j += 1
// Situation one: the current is equal to the previous ( util the last )
}
else{
temp.append(String(j))
temp.append(ans[i-1])
// Situation two: the current is not equal to the previous ( util the last )
j = 1
}
}
// Situation three: the last value
temp.append(String(j))
temp.append(ans.last!)
ans = temp
print(ans.reduce("", +))
}
return ans.reduce("", +)
}
}
With Recursion, which is a little leaner.
class Solution {
func countAndSay(_ n: Int) -> String {
if n == 1 {
return "1"
}
let str = countAndSay(n-1)
let arr = Array(str).map{Int(String($0))}
var last = arr[0]!
var temp : [(count: Int, val: Int)] = [(1, last)]
for num in arr[1...]{
if last == num{
var final = temp.last!
final.count += 1
temp.removeLast()
temp.append(final)
}
else{
temp += [(1, num!)]
}
last = num!
}
var result = ""
for key in temp{
result += "(key.0)(key.1)"
}
return result
}
}
How to go on doing some improvement?
programming-challenge swift
asked 11 mins ago
dengAprodengApro
176111
$endgroup$
add a comment |
$begingroup$
38. Count and Say
The count-and-say sequence is the sequence of integers with the first five terms as following:
1. 1
2. 11
3. 21
4. 1211
5. 111221
1is read off as"one 1"or11.
11is read off as"two 1s"or21.
21is read off as"one 2, then one 1"or1211.
Given an integer n where 1 ≤ n ≤ 30, generate the nth term of the count-and-say sequence.
Note: Each term of the sequence of integers will be represented as a string.
With iteration:
How to reduce 3 kinds of situation to one solution?
class Solution {
func countAndSay(_ n: Int) -> String {
var ans = ["1"]
var temp: [String], j: Int
for _ in 1..<n{
// from the beginning 1 to what we what
j = 1
temp =
for i in 1..<ans.count{
if ans[i - 1] == ans[i]{
j += 1
// Situation one: the current is equal to the previous ( util the last )
}
else{
temp.append(String(j))
temp.append(ans[i-1])
// Situation two: the current is not equal to the previous ( util the last )
j = 1
}
}
// Situation three: the last value
temp.append(String(j))
temp.append(ans.last!)
ans = temp
print(ans.reduce("", +))
}
return ans.reduce("", +)
}
}
With Recursion, which is a little leaner.
class Solution {
func countAndSay(_ n: Int) -> String {
if n == 1 {
return "1"
}
let str = countAndSay(n-1)
let arr = Array(str).map{Int(String($0))}
var last = arr[0]!
var temp : [(count: Int, val: Int)] = [(1, last)]
for num in arr[1...]{
if last == num{
var final = temp.last!
final.count += 1
temp.removeLast()
temp.append(final)
}
else{
temp += [(1, num!)]
}
last = num!
}
var result = ""
for key in temp{
result += "(key.0)(key.1)"
}
return result
}
}
How to go on doing some improvement?
programming-challenge swift
asked 11 mins ago
dengAprodengApro
176111
$endgroup$
38. Count and Say
The count-and-say sequence is the sequence of integers with the first five terms as following:
1. 1
2. 11
3. 21
4. 1211
5. 111221
1is read off as"one 1"or11.
11is read off as"two 1s"or21.
21is read off as"one 2, then one 1"or1211.
Given an integer n where 1 ≤ n ≤ 30, generate the nth term of the count-and-say sequence.
Note: Each term of the sequence of integers will be represented as a string.
With iteration:
How to reduce 3 kinds of situation to one solution?
class Solution {
func countAndSay(_ n: Int) -> String {
var ans = ["1"]
var temp: [String], j: Int
for _ in 1..<n{
// from the beginning 1 to what we what
j = 1
temp =
for i in 1..<ans.count{
if ans[i - 1] == ans[i]{
j += 1
// Situation one: the current is equal to the previous ( util the last )
}
else{
temp.append(String(j))
temp.append(ans[i-1])
// Situation two: the current is not equal to the previous ( util the last )
j = 1
}
}
// Situation three: the last value
temp.append(String(j))
temp.append(ans.last!)
ans = temp
print(ans.reduce("", +))
}
return ans.reduce("", +)
}
}
With Recursion, which is a little leaner.
class Solution {
func countAndSay(_ n: Int) -> String {
if n == 1 {
return "1"
}
let str = countAndSay(n-1)
let arr = Array(str).map{Int(String($0))}
var last = arr[0]!
var temp : [(count: Int, val: Int)] = [(1, last)]
for num in arr[1...]{
if last == num{
var final = temp.last!
final.count += 1
temp.removeLast()
temp.append(final)
}
else{
temp += [(1, num!)]
}
last = num!
}
var result = ""
for key in temp{
result += "(key.0)(key.1)"
}
return result
}
}
How to go on doing some improvement?
programming-challenge swift
programming-challenge swift
asked 11 mins ago
dengAprodengApro
176111
asked 11 mins ago
dengAprodengApro
176111
asked 11 mins ago
dengAprodengApro
176111
asked 11 mins ago
dengAprodengApro
176111
asked 11 mins ago
dengAprodengApro
176111
176111
add a comment |
add a comment |
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