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 Strings, Count and Say, Design Compressed String Iterator.