LeetCode443 - String Compression

题设

Given an array of characters chars, compress it using the following algorithm:

Begin with an empty string s. For each group of consecutive repeating characters in chars:

• If the group’s length is 1, append the character to s.
• Otherwise, append the character followed by the group’s length.

The compressed string s should not be returned separately, but instead, be stored in the input character array chars. Note that group lengths that are 10 or longer will be split into multiple characters in chars.

After you are done modifying the input array, return the new length of the array.

You must write an algorithm that uses only constant extra space.

Example 1:

Input: chars = ["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: The groups are "aa", "bb", and "ccc". This compresses to "a2b2c3".

Example 2:

Input: chars = ["a"]
Output: Return 1, and the first character of the input array should be: ["a"]
Explanation: The only group is "a", which remains uncompressed since it's a single character.

Example 3:

Input: chars = ["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: The groups are "a" and "bbbbbbbbbbbb". This compresses to "ab12".

Constraints:

• 1 <= chars.length <= 2000
• chars[i] is a lowercase English letter, uppercase English letter, digit, or symbol.

题解

数据结构课作业，我直接贴上leetcode官网版本了。一道非常简单的双指针？用两个指针分别记录新数组的下一个位置和旧数组的下一个位置即可。

class Solution {
 public:
  int compress(vector<char>& chars) {
    char pre = chars[0];
    int k = 1, j = 0;
    int s = chars.size();
    for (int i = 1; i < s; i++) {
      if (chars[i] == pre)
        k++;
      else {
        chars[j++] = pre;
        pre = chars[i];
        /* k的每一位都要作为char填入新数组 */
        if (k > 1) {
          string k_str = to_string(k);
          for (char ch : k_str) chars[j++] = ch;
        }
        k = 1;
      }
    }
    chars[j++] = pre;
    if (k > 1) {
      string k_str = to_string(k);
      for (char ch : k_str) {
        chars[j++] = ch;
      }
    }
    return j;
  }
};