LeetCode：String Compression（压缩字符串）

题目

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?

Example1 :

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".

Example2 :

Input:
["a"]

Output:
Return 1, and the first 1 characters of the input array should be: ["a"]

Explanation:
Nothing is replaced.

Example3:

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.

思路

整体思路：从头开始，比对前后字符是否相同，如果相同，就删除后边的字符，计数+1；如果不同，判断前一个字符的计数值，如果大于1就在后一个字符前插入计数值。

该题考察是对矢量类型的删除和插入操作，分别对应erase和insert函数。

需要注意的是，对于删除操作，迭代器指向的地址会被删除，而erase会返回被删除元素的后一个元素地址，因此利用erase返回值给迭代器更新地址：Iter=chars.erase(Iter);。

这里还设置了一个考点是将十进制数按位以char类型存储，因此在插入计数值时，需要对10取余按位插入同时需要记录插入的位数，最后让迭代器移动对应位数，回到插入前对应的位置。

代码

class Solution {
public:
    int compress(vector<char>& chars) {
        int n=chars.size();
        if(n==0||n==1)
            return n;
        vector <char>::iterator Iter;  
        int m=0;
        char temp;
        for(Iter=chars.begin();Iter!=chars.end();)
        {
            if(m==0)
            {
                temp=*Iter;
                m++;
            }
            else
            {
                if(temp==*Iter)
                {
                    m++;
                    Iter=chars.erase(Iter);
                }
                else
                {
                    if(m==1)
                    {
                        temp=*Iter;
                        Iter++;
                        continue;
                    }
                    while(m%10||m/10)
                    {
                        chars.insert(Iter,m%10+'0');
                        m/=10;      
                    }
                    Iter++;
                }           
                continue;
            }
            Iter++;
        }


        if(m>1)
        {
            Iter=chars.end();
            while(m%10||m/10)
            {
                chars.insert(Iter,m%10+'0');
                m/=10;
                Iter=chars.end()-1;
            }
        }

        return chars.size();
    }
};