[LeetCode] 443. String Compression

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.

字符串压缩。

题意是给一个字符串，请按规则压缩。压缩的结果还是存在char array中但是最后输出的是新的char array的长度。

基本逻辑是要求你把连续出现多个的字母转化成字母 + 出现次数的组合。要求是不使用额外空间。这个题我也不知道该归为什么算法或者方法，我这里解释一下我的思路吧。做一个while循环和一个指针，首先确保指针不能跳出while循环，条件自然是指针指向的index不能大于input字符串的长度。在遍历input的时候，如果遇到一个新的字母c，先记下这个字母，然后往后遍历看看这个字母是否有连续出现，若有，则需要记录连续出现的次数count。当连续出现的字符停止之后，此时可以试着将已经遍历过的部分写入res，因为你有了c和他的count，所以就可以写入res了。注意需要将count先转化成String.valueOf然后再转化成charArray，才能将比如12转化成"1", "2"这样的形式。只要count不等于1，那么这个数字就一定要被写入最后的char array。

时间O(n)

空间O(1) - 题目要求

Java实现

class Solution {
    public int compress(char[] chars) {
        int res = 0;
        int index = 0;
        while (index < chars.length) {
            char cur = chars[index];
            int count = 0;
            while (index < chars.length && chars[index] == cur) {
                index++;
                count++;
            }
            // write the letter
            chars[res++] = cur;
            // write the occurance
            if (count != 1) {
                for (char c : String.valueOf(count).toCharArray()) {
                    chars[res++] = c;
                }
            }
        }
        return res;
    }
}

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