作者:狗且偷生 | 来源:互联网 | 2023-05-17 12:13
原题链接在这里:https:leetcode.comproblemsstring-compression题目:Givenanarrayofcharacters,compressit
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 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Design Compressed String Iterator.