共计 1523 个字符，预计需要花费 4 分钟才能阅读完成。
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;
}
