In a string S of lowercase letters, these letters form consecutive groups of the same character.
For example, a string like S = "abbxxxxzyy"has the groups "a", "bb", "xxxx", "z"and "yy".
Call a group large if it has 3 or more characters. We would like the starting and ending positions of every large group.
The final answer should be in lexicographic order.
Example 1:
Input: "abbxxxxzzy"
Output: [[3,6]]
Explanation: "xxxx" is the single large group with starting 3 and ending positions 6.
Example 2:
Input: "abc" Output: [] Explanation: We have "a","b" and "c" but no large group.
Example 3:
Input: "abcdddeeeeaabbbcd" Output: [[3,5],[6,9],[12,14]]
Note: 1 <= S.length <= 1000
在一个由小写字母构成的字符串 S 中，包含由一些连续的相同字符所构成的分组。
例如，在字符串 S = "abbxxxxzyy" 中，就含有 "a", "bb", "xxxx", "z" 和 "yy" 这样的一些分组。
我们称所有包含大于或等于三个连续字符的分组为较大分组。找到每一个较大分组的起始和终止位置。
最终结果按照字典顺序输出。
示例 1:
输入: "abbxxxxzzy"
输出: [[3,6]]
解释: "xxxx" 是一个起始于 3 且终止于 6 的较大分组。
示例 2:
输入: "abc" 输出: [] 解释: "a","b" 和 "c" 均不是符合要求的较大分组。
示例 3:
输入: "abcdddeeeeaabbbcd" 输出: [[3,5],[6,9],[12,14]]
说明: 1 <= S.length <= 1000
16ms
1 class Solution {
2 func largeGroupPositions(_ S: String) -> [[Int]] {
3 var left = 0
4 let chars = Array(S)
5 var result = [[Int]]()
6 for i in 0..<chars.count {
7 if chars[left] == chars[i] {
8 continue
9 } else {
10 if i - left >= 3 {
11 result.append([left, i - 1])
12 }
13 left = i
14 }
15 }
16 if chars.count - left >= 3 {
17 result.append([left, chars.count - 1])
18 }
19 return result
20 }
21 }
20ms
1 class Solution {
2 func largeGroupPositions(_ S: String) -> [[Int]] {
3 let chars = S.utf8CString
4 var char = chars[0]
5 var count = 1,start = 0
6 var location: [[Int]] = []
7 for i in 1..<chars.count {
8 if char == chars[i] {
9 count += 1
10 } else {
11 if count >= 3 {
12 location.append([start, start+count-1])
13 }
14 char = chars[i]
15 start = i
16 count = 1
17 }
18 }
19 return location
20 }
21 }
20ms
1 class Solution {
2 func largeGroupPositions(_ S: String) -> [[Int]] {
3 var result = [[Int]]()
4 var array = Array(S)
5
6 if array.count <= 2 {
7 return result
8 }
9
10 var i = 2
11 while i < array.count {
12 var start = -1
13 if array[i] == array[i - 1] && array[i - 1] == array[i - 2] {
14 start = i - 2
15 }
16 if start == -1 {
17 i += 1
18 }
19 else {
20 i += 1
21 while i < array.count && array[i] == array[start] {
22 i += 1
23 }
24 result.append([start, i - 1])
25 }
26 }
27 return result
28 }
29 }
24ms
1 class Solution {
2 func largeGroupPositions(_ S: String) -> [[Int]] {
3 let A = Array(S)
4 var mem: [[Int]] = []
5 var begin = 0
6 var element = A[0]
7 for (index, item) in A.enumerated() {
8 if(item == element){
9 continue
10 } else {
11 let gap = index-begin
12 if(gap>=3){
13 let temp = [begin, index-1]
14 mem.append(temp)
15 }
16 begin = index
17 element = item
18 }
19 }
20
21 if(begin != -1 && A.count - begin >= 3){
22 mem.append([begin, A.count-1])
23 }
24 return mem
25 }
26 }
36ms
1 class Solution {
2 func largeGroupPositions(_ S: String) -> [[Int]] {
3 var chars = Array(S)
4 let count = S.count
5 var res = [[Int]]()
6 var i = 0
7
8 for j in 0..<count {
9 if j == count - 1 || chars[j] != chars[j + 1] {
10 if j - i + 1 >= 3 {
11 res.append([i, j])
12 }
13 i = j + 1
14 }
15 }
16
17 return res
18 }
19 }
48ms
1 class Solution {
2 func largeGroupPositions(_ S: String) -> [[Int]] {
3 var currentStart = 0
4 var currentEnd = 0
5 var currentChar: Character = "\n"
6 var result = [[Int]]()
7 for (index,char) in S.enumerated() {
8 if char != currentChar {
9 if currentEnd - currentStart >= 2 {
10 result.append([currentStart,currentEnd])
11 }
12 currentStart = index
13 currentEnd = index
14 currentChar = char
15 } else {
16 currentEnd = index
17 }
18 }
19 if currentEnd - currentStart >= 2 {
20 result.append([currentStart,currentEnd])
21 }
22 return result
23 }
24 }