leetcode443. String Compression

題目要求

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.

對字符串進行簡單的壓縮操做，壓縮的規則是，若是出現多個重複的字母，則用字母加上字母出現的字數進行表示。若是字母只出現一次，則不記錄次數。

思路和代碼

核心思路是用三個指針分別記錄三個下標：
p1: 記錄壓縮後的內容的插入下標
p2: 記錄當前相同字符串的起始位置
p3: 記錄當前和起始位置比較的字符串的位置

一旦出現p3的值不等於p2或是p3的值大於字符數組的長度，則將壓縮結果從p1開始填寫，實現O(1)的空間複雜度。

public int compress(char[] chars) {
        int p1 = 0;
        int p2 = 0;
        int p3 = 1;
        while(p2 < chars.length) {
            if(p3 >= chars.length || chars[p3] != chars[p2]) {
                int length = p3 - p2;
                chars[p1++] = chars[p2];
                if(length != 1) {
                    int count = 0;
                    while(length != 0) {
                        int num = length % 10;
                        for(int i = p1+count ; i>p1 ; i--) {
                            chars[i] = chars[i-1];
                        }
                        chars[p1] = (char)('0' + num);
                        length /= 10;
                        count++;
                    }
                    p1 += count;
                }
                p2 = p3;
            }
            p3++;

        }
        return p1;
    }