443. String Compression

问题描述

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:

I**nput:**
[“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.

题目链接：

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思路分析

给一个char的数组，其中有重复的数字，将这个string进行压缩，如果字符连续出现的次数大于一次，就用char+count的全字符形式来表示。要求用常数空间in-place完成。最后返回新string的长度。

使用快慢两个指针来遍历数组。慢指针current指向现在所在的字符，快指针i用来进行遍历，找应该放在current位的字符。

我们需要进行压缩的条件是，到达了数组最后一个元素或者i与i+1不同了。此时我们将current更新为i的值，然后判断cout的次数，如果大于1，那么就将count变为string后插入到current之后到位置。最后将count清零。

代码

class Solution {
public:
    int compress(vector<char>& chars) {
        int current = 0, count = 0;
        for(int i = 0; i < chars.size(); i++){
            count++;
            if(i == chars.size() - 1 || chars[i] != chars[i + 1]){
                chars[current++] = chars[i];
                if(count != 1){
                    string temp = to_string(count);
                    for(char ch : temp){
                        chars[current++] = ch;
                    }
                }
                count = 0;
            }
        }
        return --current;
    }
}; 

时间复杂度：O(n)$O(n)$
空间复杂度：O(1)$O(1)$

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反思

对于current指针的移动，要多加注意，我们可以理解为current处是空的：首先是出现不同字符时，将新的字符放在current后，current就可以移动了；然后时count转换为char时，也是放一个动一个。

同样的，对于count，正常情况下跟随icount，使用i == i+1的判断，可以帮助count简化情况。