原題鏈接在這里：https://leetcode.com/problems/string-compression/description/

題目：

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.

題解：

計數連續相同char的個數加在后面.

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 mark = 0; 8 int write = 0; 9 for(int i = 0; i<chars.length; i++){ 10 if(i+1==chars.length || chars[i]!=chars[i+1]){ 11 chars[write++] = chars[mark]; 12 if(mark<i){ 13 for(char c : (""+(i-mark+1)).toCharArray()){ 14 chars[write++] = c; 15 } 16 } 17 mark = i+1; 18 } 19 } 20 21 return write; 22 } 23 }

?

轉載于:https://www.cnblogs.com/Dylan-Java-NYC/p/8202915.html

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