443 String Compression 压缩字符串
Description:
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:
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.
题目描述:
给定一组字符，使用原地算法将其压缩。
压缩后的长度必须始终小于或等于原数组长度。
数组的每个元素应该是长度为1 的字符（不是 int 整数类型）。
在完成原地修改输入数组后，返回数组的新长度。
进阶：
你能否仅使用O(1) 空间解决问题？
示例：
示例 1：
输入：
["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"]
说明：
"aa"被"a2"替代。"bb"被"b2"替代。"ccc"被"c3"替代。
示例 2：
输入：
["a"]
输出：
返回1，输入数组的前1个字符应该是：["a"]
说明：
没有任何字符串被替代。
示例 3：
输入：
["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"]。
说明：
由于字符"a"不重复，所以不会被压缩。"bbbbbbbbbbbb"被“b12”替代。
注意每个数字在数组中都有它自己的位置。
注意：
所有字符都有一个ASCII值在[35, 126]区间内。
1 <= len(chars) <= 1000。
思路:
使用一个长度指针记录返回结果, 并记录重复的值的长度
时间复杂度O(n), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
int compress(vector<char>& chars)
{
int i = 0, j = 0, l = chars.size();
while (i < l && j < l)
{
chars[i++] = chars[j];
int temp = j;
while (j < l && chars[i - 1] == chars[j]) j++;
if (j - temp > 1)
{
for (char c : to_string(j - temp)) chars[i++] = c;
}
}
return i;
}
};
Java:
class Solution {
public int compress(char[] chars) {
int i = 0, j = 0, l = chars.length;
while (i < l && j < l) {
chars[i++] = chars[j];
int temp = j;
while (j < l && chars[i - 1] == chars[j]) j++;
if (j - temp > 1) {
for (char c : String.valueOf(j - temp).toCharArray()) chars[i++] = c;
}
}
return i;
}
}
Python:
class Solution:
def compress(self, chars: List[str]) -> int:
i, j, l = 0, 0, len(chars)
while i < l and j < l:
chars[i] = chars[j]
i += 1
temp = j
while j < l and chars[i - 1] == chars[j]:
j += 1
if j - temp <= 1 :
continue
for c in str(j - temp):
chars[i] = c
i += 1
return i
