leetcode443

题目要求

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.

对字符串进行简单的压缩操作，压缩的规则是，如果出现多个重复的字母，则用字母加上字母出现的字数进行表示。如果字母只出现一次，则不记录次数。

思路和代码

核心思路是用三个指针分别记录三个下标：
p1: 记录压缩后的内容的插入下标
p2: 记录当前相同字符串的起始位置
p3: 记录当前和起始位置比较的字符串的位置

一旦出现p3的值不等于p2或是p3的值大于字符数组的长度，则将压缩结果从p1开始填写，实现O(1)的空间复杂度。

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;
}