[Leetcode] 443. String Compression 解题报告

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本文介绍了一种针对字符数组的原地压缩算法实现。该算法遍历输入数组，遇到重复字符时进行计数，非重复字符直接保留，重复字符用字符加计数的方式表示。文章提供了详细的算法思路及C++代码实现。

题目：

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.

思路：

Easy级别的题目做起来就是爽：遍历数组元素，如果发现和当前元素一样，就增加计数；否则就修改字符串中的元素，并且在个数大于1的时候同时写入计数。值得注意的是在最后还需要检查一下计数是不是大于1，如果大于1则还要额外再写入计数。

代码：

class Solution {
public:
    int compress(vector<char>& chars) {
        if (chars.size() == 0) {
            return 0;
        }
        char c = chars[0];
        int ret = 1, count = 1;
        for (int i = 1; i < chars.size(); ++i) {
            if (chars[i] == c) {
                ++count;
            }
            else {
                if (count > 1) {
                    string s = to_string(count);
                    for (auto ch : s) {
                        chars[ret++] = ch;
                    }
                }
                c = chars[i], count = 1;
                chars[ret++] = c;
            }
        }
        if (count > 1) {
            string s = to_string(count);
            for (auto ch : s) {
                chars[ret++] = ch;
            }
        }
        chars.resize(ret);
        return ret;
    }
};