LeetCode 443. String Compression

原题链接在这里:https://leetcode.com/problems/string-compression/

题目:

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.

题解:

Accumlate the count of repeating chars, if count > 1, append to the char.

Use (""+count).toCharArray() to easily append int as char array. 

Time Complexity: O(chars.length). Space: O(1).

AC Java:

 1 class Solution {
 2     public int compress(char[] chars) {
 3         if(chars == null || chars.length == 0){
 4             return 0;
 5         }
 6         
 7         int pos = 0;
 8         int count = 0;
 9         int i = 0;
10         while(i<chars.length){
11             char cur = chars[i];
12             while(i<chars.length && chars[i] == cur){
13                 count++;
14                 i++;
15             }
16             
17             chars[pos++] = chars[i-1];
18             if(count > 1){
19                 for(char c : (""+count).toCharArray()){
20                     chars[pos++] = c;
21                 }
22             }
23 
24             count = 0;
25         }
26         
27         return pos;
28     }
29 }

类似Encode and Decode StringsCount and SayDesign Compressed String Iterator.

原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/8202915.html