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➤微信公衆号：山青詠芝（shanqingyongzhi）
➤部落格園位址：山青詠芝（https://www.cnblogs.com/strengthen/）
➤GitHub位址：https://github.com/strengthen/LeetCode
➤原文位址：https://www.cnblogs.com/strengthen/p/9824714.html
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Your friend is typing his
name
into a keyboard. Sometimes, when typing a character
c
, the key might get long pressed, and the character will be typed 1 or more times.
You examine the
typed
characters of the keyboard. Return
True
if it is possible that it was your friends name, with some characters (possibly none) being long pressed.
Example 1:
Input: name = "alex", typed = "aaleex"
Output: true
Explanation: 'a' and 'e' in 'alex' were long pressed.
Example 2:
Input: name = "saeed", typed = "ssaaedd"
Output: false
Explanation: 'e' must have been pressed twice, but it wasn't in the typed output.
Example 3:
Input: name = "leelee", typed = "lleeelee"
Output: true
Example 4:
Input: name = "laiden", typed = "laiden"
Output: true
Explanation: It's not necessary to long press any character.
Note:
-
name.length <= 1000
-
typed.length <= 1000
- The characters of
name
typed
你的朋友正在使用鍵盤輸入他的名字
name
。偶爾，在鍵入字元
c
時，按鍵可能會被長按，而字元可能被輸入 1 次或多次。
你将會檢查鍵盤輸入的字元
typed
。如果它對應的可能是你的朋友的名字（其中一些字元可能被長按），那麼就傳回
True
。
示例 1：
輸入：name = "alex", typed = "aaleex"
輸出：true
解釋：'alex' 中的 'a' 和 'e' 被長按。
示例 2：
輸入：name = "saeed", typed = "ssaaedd"
輸出：false
解釋：'e' 一定需要被鍵入兩次，但在 typed 的輸出中不是這樣。
示例 3：
輸入：name = "leelee", typed = "lleeelee"
輸出：true
示例 4：
輸入：name = "laiden", typed = "laiden"
輸出：true
解釋：長按名字中的字元并不是必要的。
提示：
-
name.length <= 1000
-
typed.length <= 1000
-
name
typed
8ms
1 class Solution {
2 func isLongPressedName(_ name: String, _ typed: String) -> Bool {
3 guard name.count != 0 else {
4 return true
5 }
6 let chars1 = Array(name)
7 let chars2 = Array(typed)
8 var index = 0
9 for i in 0..<chars2.count {
10 if index == chars1.count { // Best Idea
11 if chars2[i] != chars2[i - 1] {
12 return false
13 }
14 continue
15 }
16
17 if chars1[index] == chars2[i] {
18 index += 1
19 } else if i != 0 && chars2[i] == chars2[i - 1] {
20 continue
21 } else {
22 return false
23 }
24 }
25 return index == chars1.count
26 }
27 } 12ms
1 class Solution {
2 func count(_ str: String) -> [(Character, Int)] {
3 var result = [(Character,Int)]()
4
5 var pos = str.startIndex
6 while pos < str.endIndex {
7 var next = str.index(after:pos)
8 var c = 1
9 while next < str.endIndex && str[pos] == str[next] {
10 next = str.index(after:next)
11 c += 1
12 }
13 result.append((str[pos], c))
14 pos = next
15 }
16
17
18 return result
19 }
20
21
22 func isLongPressedName(_ name: String, _ typed: String) -> Bool {
23 let nameC = count(name)
24 let typedC = count(typed)
25
26 if nameC.count != typedC.count {
27 return false
28 }
29
30 for var i in 0..<nameC.count {
31 if nameC[i].0 != typedC[i].0 {
32 return false
33 }
34 if nameC[i].1 > typedC[i].1 {
35 return false
36 }
37 }
38
39 return true
40 }
41 } 16ms
1 class Solution {
2 func isLongPressedName(_ name: String, _ typed: String) -> Bool {
3 guard name.count != 0 else {
4 return true
5 }
6 let chars1 = Array(name)
7 let chars2 = Array(typed)
8 var index = 0
9 for i in 0..<chars2.count {
10 if index == chars1.count { // Best Idea
11 if chars2[i] != chars2[i - 1] {
12 return false
13 }
14 continue
15 }
16
17 if chars1[index] == chars2[i] {
18 index += 1
19 } else if i != 0 && chars2[i] == chars2[i - 1] {
20 continue
21 } else {
22 return false
23 }
24 }
25 return index == chars1.count
26 }
27 } 36ms
1 class Solution {
2 func isLongPressedName(_ name: String, _ typed: String) -> Bool {
3 guard name.count != 0 else {
4 return true
5 }
6 var index1 = 0
7 var index2 = 0
8 var char1 = Array(name)
9 var char2 = Array(typed)
10 while index1 < name.count && index2 < typed.count {
11 if char1[index1] == char2[index2] {
12 index1 += 1
13 index2 += 1
14 } else if index2 != 0 && char2[index2] == char2[index2 - 1] {
15 index2 += 1
16 } else {
17 return false
18 }
19 }
20 return index1 == char1.count
21 }
22 } 36ms
1 class Solution {
2 func isLongPressedName(_ name: String, _ typed: String) -> Bool {
3 var p:Int = 0
4 for num in 0..<typed.count
5 {
6 var typeIndex = typed.index(typed.startIndex,offsetBy: num)
7 var nameIndex = name.index(name.startIndex,offsetBy: p)
8 if p < name.count && typed[typeIndex] == name[nameIndex]
9 {
10 p += 1
11 }
12 else
13 {
14 if num > 0 && typed[typeIndex] == typed[typed.index(typed.startIndex,offsetBy:num - 1)]
15 {
16 continue
17 }
18 else
19 {
20 return false
21 }
22 }
23 }
24 return p == name.count
25 }
26 } 44ms
1 class Solution {
2 func isLongPressedName(_ name: String, _ typed: String) -> Bool {
3 guard name.count != 0 else {
4 return true
5 }
6
7 var index1 = 0
8 var index2 = 0
9 var char1 = Array(name)
10 var char2 = Array(typed)
11 while index1 < name.count && index2 < typed.count {
12 if char1[index1] == char2[index2] {
13 index1 += 1
14 index2 += 1
15 } else {
16 if index2 != 0 && char2[index2] == char2[index2 - 1] {
17 index2 += 1
18 } else {
19 return false
20 }
21 }
22 }
23 guard index1 == name.count else {
24 return false
25 }
26
27 while index2 < typed.count {
28 if index2 != 0 && char2[index2] == char2[index2 - 1] {
29 index2 += 1
30 continue
31 } else {
32 break
33 }
34 }
35 return index2 == typed.count
36 }
37 } 52ms
1 class Solution {
2 func isLongPressedName(_ name: String, _ typed: String) -> Bool {
3 guard name.count != 0 else {
4 return true
5 }
6
7 var index1 = 0
8 var index2 = 0
9 var char1 = Array(name)
10 var char2 = Array(typed)
11 while index1 < name.count && index2 < typed.count {
12 if char1[index1] == char2[index2] {
13 index1 += 1
14 index2 += 1
15 } else if index2 != 0 && char2[index2] == char2[index2 - 1] {
16 index2 += 1
17 } else {
18 return false
19 }
20 }
21 guard index1 == name.count else {
22 return false
23 }
24
25 while index2 < typed.count {
26 if index2 != 0 && char2[index2] == char2[index2 - 1] {
27 index2 += 1
28 continue
29 } else {
30 break
31 }
32 }
33 return index2 == typed.count
34 }
35 } 轉載于:https://www.cnblogs.com/strengthen/p/9824714.html
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