leetcode-443-String Compression

题目描述：

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.

 

要完成的函数：

int compress(vector<char>& chars) 

说明：

1、看完example之后才明白这道题要干嘛。要求在原本vector上面修改，然后输出最后vector的长度。空间复杂度是O(1)。

2、这道题目不难做，就是比较综合，考察多种操作。做法如下：

从vector首位开始处理，把碰到的第一个数记录下来，后续的数逐个跟它比较，如果相等就把后续的数删去，顺便记录删了几个数。

然后如果确实删去了数，那么把删去的数的个数+1，拆成一位一位的，插入在vector中，要注意插入的位置。

如果没有删去重复的后续的数，也就是记录下来的数只出现了一次，那么继续处理下一个数，仍然记录下来，覆盖掉原本的数。

直到vector所有字符都处理完毕。

有同学可能会疑惑为什么要删除和插入vector中的元素，因为题目要求要in-place处理，而且样例也有要求原vector要处理。

代码如下：

    int compress(vector<char>& chars) 
    {
        int i=0,j,count,ret=0;
        char t1;
        while(i<chars.size())
        {
            t1=chars[i];//t1记录碰到的数
            count=1;//出现了一次
            i++;//i指向下一个要处理的数
            while(chars[i]==t1&&i<chars.size())//如果出现重复了
            {
                chars.erase(chars.begin()+i);//把这个重复的数删掉，删完之后i没有变，仍然指向第i个
                count++;//出现了多一次
            }
            if(count==1)
                ret++;
            else
            {
                ret++;
                j=i;//记录i的数值，i此时指向和上一个不重复的数
                while(count!=0)
                {
                    ret++;
                    chars.insert(chars.begin()+i,count%10+'0');
                    j++;//记录i之后应该要指向第几个数
                    count/=10;
                }
                i=j;//i仍然指向下一个要处理的数
            }
            
        }
        return ret;
    }

上述代码实测8ms，beats 82.94% of cpp submissions。