2017年12月21日 星期四

[LeetCode] 443. String Compression

轉自LeetCode

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:
  1. All characters have an ASCII value in [35, 126].
  2. 1 <= len(chars) <= 1000.
<Solution>

想法如下(參考資料)
  • 用三個指標,一個是寫的位置,一個是檢查的起始點,一個是檢查起始點之後的指標
  • 當發現沒有重複的時候,將檢查起始點的字母寫進寫的位置,然後按順序寫入重複的次數
  • 最後將檢查的起始點,更新到最後檢查的位置
code 如下

C++
class Solution {
public:
int compress(vector<char>& chars) {
const int len = chars.size();
const int limit = len - 1;
int writeIdx = 0, anchorIdx = 0;
string str;
for(int readIdx = 0; readIdx < len; readIdx++) {
if(readIdx == limit || chars[readIdx] != chars[readIdx + 1]) {
chars[writeIdx++] = chars[anchorIdx];
if(readIdx > anchorIdx) {
str = to_string(readIdx - anchorIdx + 1);
for(const auto c : str) {
chars[writeIdx++] = c;
}
}
anchorIdx = readIdx+1;
}
}
return writeIdx;
}
};

Java
class Solution {
public int compress(char[] chars) {
final int limit = chars.length - 1;
int writeIdx = 0, anchorIdx = 0;
for(int readIdx = 0; readIdx < chars.length; readIdx++) {
if(readIdx == limit || chars[readIdx] != chars[readIdx + 1]) {
chars[writeIdx++] = chars[anchorIdx];
if(readIdx > anchorIdx) {
for(char c : ("" + (readIdx - anchorIdx + 1)).toCharArray()) {
chars[writeIdx++] = c;
}
}
anchorIdx = readIdx + 1;
}
}
return writeIdx;
}
}

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