C#LeetCode刷题之#443-压缩字符串​​​​​​​(String Compression)

C#LeetCode刷题之#443-压缩字符串​​​​​​​(String Compression)

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: O(n)。请注意 Compress2 的时间复杂度不是 O(n^{2}) ,因为数组只被扫描了一遍。

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