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

客官°小女子只卖身不卖艺 2022-06-04 08:55 177阅读 0赞

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;
        }
    };