-
Notifications
You must be signed in to change notification settings - Fork 0
/
String Compression
81 lines (64 loc) · 1.91 KB
/
String Compression
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
/*
LeetCode 443:
Given an array of characters, compress it in-place.
The length after compression must always be smaller than or equal to the original array.
Every element of the array should be a character (not int) of length 1.
After you are done modifying the input array in-place, return the new length of the array.
Follow up:
Could you solve it using only O(1) extra space?
Example 1:
Input:
["a","a","b","b","c","c","c"]
Output:
Return 6, and the first 6 characters of the input array should be: ["a","2","b","2","c","3"]
Explanation:
"aa" is replaced by "a2". "bb" is replaced by "b2". "ccc" is replaced by "c3".
Example 2:
Input:
["a"]
Output:
Return 1, and the first 1 characters of the input array should be: ["a"]
Explanation:
Nothing is replaced.
Example 3:
Input:
["a","b","b","b","b","b","b","b","b","b","b","b","b"]
Output:
Return 4, and the first 4 characters of the input array should be: ["a","b","1","2"].
Explanation:
Since the character "a" does not repeat, it is not compressed. "bbbbbbbbbbbb" is replaced by "b12".
Notice each digit has it's own entry in the array.
Note:
All characters have an ASCII value in [35, 126].
1 <= len(chars) <= 1000.
Solution:
Language: C++
Tag: 2Pointer
*/
class Solution {
public:
int compress(vector<char>& chars) {
// 1. O(n) time + O(1) space
int res = 0, idx = 0;
int size = chars.size();
int pos = 0;
while (idx < size) {
chars[pos++] = chars[idx];
int i = idx;
while (i < size && chars[i] == chars[idx]) {
++i;
}
int count = i - idx;
++res;
if (count > 1) {
string tmp = to_string(count);
res += tmp.length();
for (int j = 0; j < tmp.length(); ++j) {
chars[pos++] = tmp[j];
}
}
idx = i;
}
return res;
}
};