443. String Compression - Easy
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.
Note:
- All characters have an ASCII value in
[35, 126]
. 1 <= len(chars) <= 1000
.
双指针+一个常数用来计数
时间:O(N),空间:O(1)
class Solution { public int compress(char[] chars) { int resIdx = 0, idx = 0; while(idx < chars.length) { char cur = chars[idx]; int cnt = 0; while(idx < chars.length && cur == chars[idx]) { idx++; cnt++; } chars[resIdx++] = cur; if(cnt != 1) { for(char c : Integer.toString(cnt).toCharArray()) { chars[resIdx++] = c; } } } return resIdx; } }
二刷:
class Solution { public int compress(char[] chars) { if(chars == null || chars.length == 0) { return 0; } int s = 0, f = 0; while(f < chars.length) { int f_begin = f; while(f < chars.length && chars[f_begin] == chars[f]) { f++; } int cnt = f - f_begin; if(cnt > 1) { chars[s++] = chars[f_begin]; String str = Integer.toString(cnt); for(int i = 0; i < str.length(); i++) { chars[s++] = str.charAt(i); } } else { chars[s++] = chars[f_begin]; } } return s; } }