[Leetcode] 443. String Compression 解题报告

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:

1. All characters have an ASCII value in [35, 126].
2. 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;    }};