zoukankan      html  css  js  c++  java
  • 443. String Compression

    package LeetCode_443
    
    /**
     * 443. String Compression
     * https://leetcode.com/problems/string-compression/
     * Given an array of characters chars, compress it using the following algorithm:
    Begin with an empty string s. For each group of consecutive repeating characters in chars:
    If the group's length is 1, append the character to s.
    Otherwise, append the character followed by the group's length.
    The compressed string s should not be returned separately, but instead be stored in the input character array chars.
    Note that group lengths that are 10 or longer will be split into multiple characters in chars.
    After you are done modifying the input array, return the new length of the array.
    
    Follow up:
    Could you solve it using only O(1) extra space?
    
    Example 1:
    Input: chars = ["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: The groups are "aa", "bb", and "ccc". This compresses to "a2b2c3".
    
    Example 2:
    Input: chars = ["a"]
    Output: Return 1, and the first character of the input array should be: ["a"]
    Explanation: The only group is "a", which remains uncompressed since it's a single character.
    
    Example 3:
    Input: chars = ["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: The groups are "a" and "bbbbbbbbbbbb". This compresses to "ab12".
    
    Constraints:
    1. 1 <= chars.length <= 2000
    2. chars[i] is a lower-case English letter, upper-case English letter, digit, or symbol.
     * */
    class Solution {
        /*
        * solution: scan array to counting number of char, Time:O(n), Space:O(1)
        * */
        fun compress(chars: CharArray): Int {
            var count = 1
            var index = 0
            val n = chars.size
            var i = 0
            while (i < n) {
                //check if current char equals next one,if yes, increasing i and count
                while (i < n - 1 && chars[i] == chars[i + 1]) {
                    count++
                    i++
                }
                if (count == 1) {
                    //count==1, just replace char
                    chars[index++] = chars[i]
                } else {
                    //count>1, replace char and count
                    chars[index++] = chars[i]
                    //handle the number
                    val countStr = count.toString()
                    for (k in countStr.indices) {
                        chars[index++] = countStr[k]
                    }
                }
                i++
                //reset count
                count = 1
            }
            return index
        }
    }
  • 相关阅读:
    鲲鹏服务器测试
    缓存区溢出实验
    读书笔记
    《信息安全系统设计与实现》学习笔记9
    改进ls的实现
    团队作业(四):描述设计
    《需求规格书》修订版
    反汇编测试
    《信息安全系统设计与实现》学习笔记8
    stat命令的实现-mystat
  • 原文地址:https://www.cnblogs.com/johnnyzhao/p/14279239.html
Copyright © 2011-2022 走看看