标签：pre span algorithm tput Plan one tps example ext
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".
class Solution { public int compress(char[] chars) { if (chars.length <= 1) { return chars.length; } int index = 0; int res = 0; while (index < chars.length) { int curCount = 0; char curChar = chars[index]; while (index < chars.length && chars[index] == curChar) { index += 1; curCount += 1; } chars[res++] = curChar; if (curCount != 1) { for (char c: Integer.toString(curCount).toCharArray()) { chars[res] = c; res += 1; } } } return res; } }
标签：pre span algorithm tput Plan one tps example ext
原文地址：https://www.cnblogs.com/xuanlu/p/12182187.html
