LeetCode 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:
class Solution {
public int compress(char[] chars) {
if(chars == null || chars.length == 0){
return 0;
}
int pos = 0;
int count = 0;
int i = 0;
while(i<chars.length){
char cur = chars[i];
while(i<chars.length && chars[i] == cur){
count++;
i++;
}
chars[pos++] = chars[i-1];
if(count > 1){
for(char c : (""+count).toCharArray()){
chars[pos++] = c;
}
}
count = 0;
}
return pos;
}
}
类似Encode and Decode Strings, Count and Say, Design Compressed String Iterator.
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