leetcode443. String Compression

nifhlheimr 发布于2019-08-16 16:41 / 1212人阅读

摘要：题目要求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。

题目要求

</>复制代码 
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。

</>复制代码 
    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;
    }

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