[LeetCode] 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.

 

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:

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

将字符串中的字符进行压缩，如果某字符只出现了一次，则不用数字表示该数字的次数。每个字符出现次数超过1位数，则由多个1位数字表示。

思路：首先顺序统计出现每个单词出现的次数，用一个变量cur作为新修改数组的索引，指向当前被修改的字符。然后利用ostringstream字符串流来逐位输入字符出现的次数。最后返回修改的字符个数cur即可。

class Solution {
public:
    int compress(vector<char>& chars) {
        int cur = 0;
        for (int i = 1, cnt = 1; i <= chars.size(); i++) {
            if (i != chars.size() && chars[i] == chars[i - 1]) {
                cnt++;
            }
            else {
                chars[cur++] = chars[i - 1];
                if (cnt != 1) {
                    ostringstream ss;
                    ss << cnt;
                    for (auto& c : ss.str()) {
                        chars[cur++] = c;
                    }
                }
                cnt = 1;
            }
        }
        return cur;
    }
};
// 9 ms