
问题
给定一组字符,使用原地算法将其压缩。
压缩后的长度必须始终小于或等于原数组长度。
数组的每个元素应该是长度为1 的字符(不是 int 整数类型)。
在完成原地修改输入数组后,返回数组的新长度。
进阶:你能否仅使用O(1) 空间解决问题?
输入:[‘a’,’a’,’b’,’b’,’c’,’c’,’c’]
输出:返回6,输入数组的前6个字符应该是:[‘a’,’2′,’b’,’2′,’c’,’3′]
说明:“aa”被”a2″替代。”bb”被”b2″替代。”ccc”被”c3″替代。
输入:[‘a’]
输出:返回1,输入数组的前1个字符应该是:[‘a’]
说明:没有任何字符串被替代。
输入:[‘a’,’b’,’b’,’b’,’b’,’b’,’b’,’b’,’b’,’b’,’b’,’b’,’b’]
输出:返回4,输入数组的前4个字符应该是:[‘a’,’b’,’1′,’2′]。
说明:由于字符”a”不重复,所以不会被压缩。”bbbbbbbbbbbb”被“b12”替代。注意每个数字在数组中都有它自己的位置。
注意:
- 所有字符都有一个ASCII值在[35, 126]区间内。
- 1 <= len(chars) <= 1000。
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?
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”.
Input:[‘a’]
Output:Return 1, and the first 1 characters of the input array should be: [‘a’]
Explanation:Nothing is replaced.
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.
示例
public class Program { public static void Main(string[] args) { var chars = new char[] { 'a', 'b', 'b' }; var res = Compress(chars); Console.WriteLine(res); chars = new char[] { 'a', 'a', 'b', 'b', 'c', 'c', 'c' }; res = Compress2(chars); Console.WriteLine(res); Console.ReadKey(); } private static int Compress(char[] chars) { if(chars.Length == 1) return 1; var count = 1; var res = string.Empty; for(var i = 1; i < chars.Length; i++) { if(chars[i] == chars[i - 1]) { count++; } else { res += chars[i - 1].ToString() + (count == 1 ? "" : count.ToString()); count = 1; } if(i == chars.Length - 1) res += chars[i].ToString() + (count == 1 ? "" : count.ToString()); } for(int i = 0; i < res.Length; i++) { chars[i] = res[i]; } return res.Length; } private static int Compress2(char[] chars) { if(chars.Length == 1) return 1; var index = 0; var count = 1; for(var i = 1; i < chars.Length; i++) { if(chars[i] == chars[i - 1]) { count++; } else { ResetChars(chars, ref count, ref index, i - 1); } if(i == chars.Length - 1) { ResetChars(chars, ref count, ref index, i); } } return index; } private static void ResetChars(char[] chars, ref int count, ref int index, int i) { if(count != 1) { chars[index] = chars[i]; for(int j = 0; j < count.ToString().Length; j++) { chars[j + index + 1] = count.ToString()[j]; } index += count.ToString().Length + 1; } else { chars[index] = chars[i]; index++; } count = 1; } }
以上给出2种算法实现,以下是这个案例的输出结果:
3 6
分析:
显而易见,以上2种算法的时间复杂度均为: 。请注意 Compress2 的时间复杂度不是
,因为数组只被扫描了一遍。
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