给你一个字符数组 chars ，请使用下述算法压缩：
从一个空字符串 s 开始。对于 chars 中的每组 连续重复字符 ：
- 如果这一组长度为
1 ，则将字符追加到 s 中。
- 否则，需要向
s 追加字符，后跟这一组的长度。
压缩后得到的字符串 s 不应该直接返回 ，需要转储到字符数组 chars 中。需要注意的是，如果组长度为 10 或 10 以上，则在
chars 数组中会被拆分为多个字符。
请在 修改完输入数组后 ，返回该数组的新长度。
你必须设计并实现一个只使用常量额外空间的算法来解决此问题。
示例 1：
**输入：** chars = ["a","a","b","b","c","c","c"]
**输出：** 返回 6 ，输入数组的前 6 个字符应该是：["a","2","b","2","c","3"]
**解释：** "aa" 被 "a2" 替代。"bb" 被 "b2" 替代。"ccc" 被 "c3" 替代。
示例 2：
**输入：** chars = ["a"]
**输出：** 返回 1 ，输入数组的前 1 个字符应该是：["a"]
**解释：** 唯一的组是“a”，它保持未压缩，因为它是一个字符。
示例 3：
**输入：** chars = ["a","b","b","b","b","b","b","b","b","b","b","b","b"]
**输出：** 返回 4 ，输入数组的前 4 个字符应该是：["a","b","1","2"]。
**解释：** 由于字符 "a" 不重复，所以不会被压缩。"bbbbbbbbbbbb" 被 “b12” 替代。
提示：
1 <= chars.length <= 2000chars[i] 可以是小写英文字母、大写英文字母、数字或符号
方法一：双指针
思路和算法
为了实现原地压缩，我们可以使用双指针分别标志我们在字符串中读和写的位置。每次当读指针 read 移动到某一段连续相同子串的最右侧，我们就在写指针 write 处依次写入该子串对应的字符和子串长度即可。
在实际代码中，当读指针 read 位于字符串的末尾，或读指针 read 指向的字符不同于下一个字符时，我们就认为读指针 read 位于某一段连续相同子串的最右侧。该子串对应的字符即为读指针 read 指向的字符串。我们使用变量 left 记录该子串的最左侧的位置，这样子串长度即为 read} - \textit{left} + 1$。
特别地，为了达到 $O(1)$ 空间复杂度，我们需要自行实现将数字转化为字符串写入到原字符串的功能。这里我们采用短除法将子串长度倒序写入原字符串中，然后再将其反转即可。
代码
[sol1-C++]1
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| class Solution {
public:
int compress(vector<char>& chars) {
int n = chars.size();
int write = 0, left = 0;
for (int read = 0; read < n; read++) {
if (read == n - 1 || chars[read] != chars[read + 1]) {
chars[write++] = chars[read];
int num = read - left + 1;
if (num > 1) {
int anchor = write;
while (num > 0) {
chars[write++] = num % 10 + '0';
num /= 10;
}
reverse(&chars[anchor], &chars[write]);
}
left = read + 1;
}
}
return write;
}
};
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[sol1-Java]1
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| class Solution {
public int compress(char[] chars) {
int n = chars.length;
int write = 0, left = 0;
for (int read = 0; read < n; read++) {
if (read == n - 1 || chars[read] != chars[read + 1]) {
chars[write++] = chars[read];
int num = read - left + 1;
if (num > 1) {
int anchor = write;
while (num > 0) {
chars[write++] = (char) (num % 10 + '0');
num /= 10;
}
reverse(chars, anchor, write - 1);
}
left = read + 1;
}
}
return write;
}
public void reverse(char[] chars, int left, int right) {
while (left < right) {
char temp = chars[left];
chars[left] = chars[right];
chars[right] = temp;
left++;
right--;
}
}
}
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[sol1-C#]1
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| public class Solution {
public int Compress(char[] chars) {
int n = chars.Length;
int write = 0, left = 0;
for (int read = 0; read < n; read++) {
if (read == n - 1 || chars[read] != chars[read + 1]) {
chars[write++] = chars[read];
int num = read - left + 1;
if (num > 1) {
int anchor = write;
while (num > 0) {
chars[write++] = (char) (num % 10 + '0');
num /= 10;
}
Reverse(chars, anchor, write - 1);
}
left = read + 1;
}
}
return write;
}
public void Reverse(char[] chars, int left, int right) {
while (left < right) {
char temp = chars[left];
chars[left] = chars[right];
chars[right] = temp;
left++;
right--;
}
}
}
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[sol1-C]1
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| void swap(char *a, char *b) {
char t = *a;
*a = *b, *b = t;
}
void reverse(char *a, char *b) {
while (a < b) {
swap(a++, --b);
}
}
int compress(char *chars, int charsSize) {
int write = 0, left = 0;
for (int read = 0; read < charsSize; read++) {
if (read == charsSize - 1 || chars[read] != chars[read + 1]) {
chars[write++] = chars[read];
int num = read - left + 1;
if (num > 1) {
int anchor = write;
while (num > 0) {
chars[write++] = num % 10 + '0';
num /= 10;
}
reverse(&chars[anchor], &chars[write]);
}
left = read + 1;
}
}
return write;
}
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[sol1-Python3]1
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| class Solution:
def compress(self, chars: List[str]) -> int:
def reverse(left: int, right: int) -> None:
while left < right:
chars[left], chars[right] = chars[right], chars[left]
left += 1
right -= 1
n = len(chars)
write = left = 0
for read in range(n):
if read == n - 1 or chars[read] != chars[read + 1]:
chars[write] = chars[read]
write += 1
num = read - left + 1
if num > 1:
anchor = write
while num > 0:
chars[write] = str(num % 10)
write += 1
num //= 10
reverse(anchor, write - 1)
left = read + 1
return write
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[sol1-Golang]1
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| func compress(chars []byte) int {
write, left := 0, 0
for read, ch := range chars {
if read == len(chars)-1 || ch != chars[read+1] {
chars[write] = ch
write++
num := read - left + 1
if num > 1 {
anchor := write
for ; num > 0; num /= 10 {
chars[write] = '0' + byte(num%10)
write++
}
s := chars[anchor:write]
for i, n := 0, len(s); i < n/2; i++ {
s[i], s[n-1-i] = s[n-1-i], s[i]
}
}
left = read + 1
}
}
return write
}
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[sol1-JavaScript]1
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| var compress = function(chars) {
const n = chars.length;
let write = 0, left = 0;
for (let read = 0; read < n; read++) {
if (read === n - 1 || chars[read] !== chars[read + 1]) {
chars[write++] = chars[read];
let num = read - left + 1;
if (num > 1) {
const anchor = write;
while (num > 0) {
chars[write++] = '' + num % 10;
num = Math.floor(num / 10);
}
reverse(chars, anchor, write - 1);
}
left = read + 1;
}
}
return write;
};
const reverse = (chars, left, right) => {
while (left < right) {
const temp = chars[left];
chars[left] = chars[right];
chars[right] = temp;
left++;
right--;
}
}
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复杂度分析
