LeetCode String Compression 字符串压缩

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.

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 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。

题解：首先这道题目，需要用o(1)的空间复杂度来完成，那么这时候就要考虑用多个指针来满足条件，通过调用多个指针，让不同的指针指向不同的位置，即可完成。

public static int compress(char[] chars)
    {
        if(chars.length == 0)
            return 0;
        int m = 0,i = 0,j;
        while(i < chars.length)
        {
            j = i;
            chars[m++] = chars[i];
            int k = 0;
            while(j < chars.length && chars[i] == chars[j])
            {
                k++;
                j++;
            }
            StringBuilder sb = new StringBuilder();
            if(k != 1)
            {
                while(k != 0)
                {
                    sb.append((char)(k % 10 + 48));
                    k /= 10;
                }
            }
            sb.reverse();
            for(char c : sb.toString().toCharArray())
                chars[m++] = c;
            i = j;
        }
        return m;
    }

首先，需要三个指针，分别为i,j和m。其中，i指向原来数组中第一次出现某个元素的位置，而j则表示数组下标，会随着依次搜寻往后走；而m则用来记录某一个元素在数组中出现的次数。当循环条件为j不大于数组长度且第一次出现某个元素与之后的元素不一样，那么就将其结束循环，然后用一个stringbuilder来进行求个数，并反转。这里需要特别注意的是，在求元素出现个数时，要特别当心，当元素出现的次数只有1次时，不需要将其写入数组中。

LeetCode String Compression 字符串压缩

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