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《LeetCode刷题(Easy Rank):443. String Compression》

 4周前  •   LeetCode  •     •   17  •   0

Question:

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.

Example 1:

示例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:

示例Input:
["a"]

Output:
Return 1, and the first 1 characters of the input array should be: ["a"]

Explanation:
Nothing is replaced.

Example 3:

示例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.

Solution:

JAVAclass Solution {
    public int compress(char[] chars) {
        if (chars == null || chars.length == 0)
            return 0;
        int index = 0, n = chars.length, i = 0;
        while (i < n) {
            char ch = chars[i];
            int j = i;
            while (j < n && chars[i] == chars[j]) { // chars[i..j - 1] are ch.
                j++;
            }
            int freq = j - i; // The frequency of ch.
            chars[index++] = ch;
            if (freq != 1) {
                String strFreq = String.valueOf(freq); 
                for (char chFreq : strFreq.toCharArray())
                    chars[index++] = chFreq;
            } 
            i = j;
        }
        return index;
    }
}

 

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