# 443. String Compression

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.

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:

1. All characters have an ASCII value in `[35, 126]`.
2. `1 <= len(chars) <= 1000`.

``````class Solution {
public:
int compress(vector<char>& chars) {
int n = chars.size(), cur = 0;
for (int i = 0, j = 0; i < n; i = j) {
while (j < n && chars[j] == chars[i]) ++j;
chars[cur++] = chars[i];
if (j - i == 1) continue;
for (char c : to_string(j - i)) chars[cur++] = c;
}
return cur;
}
};
``````

Count and Say

Encode and Decode Strings

Design Compressed String Iterator

https://leetcode.com/problems/string-compression/discuss/118472/C++-simple-solution-(-EASY-to-understand-)-with-explanation

LeetCode All in One 题目讲解汇总(持续更新中…)

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