443. String Compression

问题描述：

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.

 

解题思路：

在现有位置上压缩，用cur来记录当前字符，count记录出现个数,若遍历到的字符：

　　1. chars[i] == cur : count++;

　　2.chars[i] != cur：将cur及其出现次数更新在chars中， 需要注意的是：当cur>1时，才把数字记入字符数组中

另有一点需要注意：

　　推出循环时，仍有最后一个字符没有计入到数组中！所以需再加一遍。

 

代码：

class Solution {
public:
    int compress(vector<char>& chars) {
        int n = chars.size();
        if(n < 2)
            return n;
        char cur = chars[0];
        int count = 0;
        int pos = 0;
        for(int i = 0; i < n; i++){
            if(chars[i] == cur){
                count++;
            }else{
                chars[pos] = cur;
                pos++;
                if(count > 1){
                    string digit = to_string(count);
                    for(int j = 0; j < digit.size(); j++)
                        chars[pos++] = digit[j];        
                }
                count = 1;
                cur = chars[i];
            }
        }
        chars[pos++] = cur;
        if(count > 1){
            string digit = to_string(count);
            for(int j = 0; j < digit.size(); j++)
                chars[pos++] = digit[j];        
        }
        return pos;
    }
};