weekly contest 56 第二题 443. String Compression

题目

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1. String Compression My SubmissionsBack to Contest
   User Accepted: 1210
   User Tried: 1446
   Total Accepted: 1223
   Total Submissions: 3403
   Difficulty: Easy
   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.

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分析

这题正常模拟

统计后直接在后一位换上结果

不明白的直接看代码

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代码

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class Solution {
public:

    void fill ( vector<char> & chars , int &pos , int num ){
        string tmp = to_string( num ) ;
        for( int i=0 ; i<tmp.size() ; i++)
            chars[pos++] = tmp[i];
    }

    int compress(vector<char>& chars) {
        if( chars.size() <= 1 )
            return chars.size() ;

        char start = chars[0] ;
        int pos = 1 ; 
        int count = 0 ;
        for( int i=0 ; i < chars.size() ; i++){
            if( chars[i] == start )
                count ++ ;
            else{

                if( count != 1 )
                    fill(chars,  pos , count );

                chars[pos++] = chars[i];

                start = chars[i] ;
                count = 1 ;
            }
        }
        if( count != 1 )
                fill(chars,  pos , count );

        return pos;

    }
};

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时间复杂度 $$O(N)$$

空间复杂度 $$O(N)$$