leetcode 443. String Compression 压缩字符串+暴力遍历

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

就是压缩字符串，使用两个指针即可完成，一个指针做遍历，另一个做结果的遍历

代码如下：

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
#include <cmath>

using namespace std;


class Solution
{
public:
    int compress(vector<char>& c)
    {
        int start = 0 , end = 0;
        int count = 0;
        for (int end = 0; end < c.size(); end++)
        {
            count++;
            if (end == c.size() - 1 || c[end] != c[end + 1])
            {
                c[start++] = c[end];
                if (count > 1)
                {
                    string tmp = to_string(count);
                    for (char t : tmp)
                        c[start++] = t;
                }
                count = 0;
            }
        }
        return start;
    }
};