超过三个连续重复的字母删除

复制代码
package com.ag;

public class Test {
    /**
     * For a given string that only contains alphabet characters a-z, if 3 or more consecutive
     * characters are identical, remove them from the string. Repeat this process until
     * there is no more than 3 identical characters sitting besides each other.
     * Example:
     * Input: aabcccbbad
     * Output:
     * -> aabbbad
     * -> aaad
     * -> d
     */
    public String removeConsecutive(String s) {
        StringBuilder result = new StringBuilder();
        if (s.length() >= 1) {
            for (int i = 0; i < s.length(); i++) {
                char currentChar = s.charAt(i);
                int count = 0;

                for (int j = 0; j < s.length(); j++) {
                    if (s.charAt(j) == currentChar) {
                        count++;
                    }
                }
                if (count <= 2) {
                    result = result.append(currentChar);
                }
            }
        }
        return result.toString();
    }


    /**
     *#Stage 2 - advanced requirement
     * Instead of removing the consecutively identical characters, replace them with a
     * single character that comes before it alphabetically.
     * Example:
     * ccc -> b
     * bbb -> a
     * Input: abcccbad
     * Output:
     * -> abbbad, ccc is replaced by b
     * -> aaad, bbb is replaced by a
     * -> d
     */
    public String replaceConsecutive(String s) {
        while (true) {
            StringBuilder newString = new StringBuilder();
            int count = 1;

            for (int i = 1; i < s.length(); i++) {
                if (i > 0 && s.charAt(i) == s.charAt(i - 1)) {
                    count++;
                } else {
                    if (count >= 3) {
                        // 替换为前一个字母
                        if ((s.charAt(i - 1) - 1) == 96) {
                            newString.append("");
                        } else {
                            char newChar = (char) (s.charAt(i - 1) - 1);
                            newString.append(newChar);
                        }
                    } else {
                        for (int j = 0; j < count; j++) {
                            newString.append(s.charAt(j));
                        }
                    }
                    count = 1;
                }
            }

            // 处理最后一组字符
            if (count >= 3) {
                char newChar = (char) (s.charAt(s.length() - 1) - 1);
                newString.append(newChar);
            } else {
                for (int j = 0; j < count; j++) {
                    newString.append(s.charAt(s.length() - 1));
                }
            }

            String newStr = newString.toString();
            if (newStr.equals(s)) { // 没有变化
                break;
            }
            s = newStr;
        }
        return s;
    }


    public static void main(String[] args) {
        Test t = new Test();
        String s1 = "aabcccbbad";
        String s2 = "abcccbad";
        System.out.println(t.removeConsecutive(s1));   // d
        System.out.println(t.replaceConsecutive(s2));  // d


    }

}
相关推荐
呆呆没有脑袋几秒前
深入浅出 JavaScript 闭包:从核心概念到框架实践
前端
snakeshe10101 分钟前
用100行代码实现React useState钩子:多状态管理揭秘
前端
爱编程的喵2 分钟前
JavaScript闭包深度解析:从作用域到实战应用
前端·javascript
写个博客24 分钟前
暑假算法日记第二天
算法
考虑考虑32 分钟前
JDK9中的dropWhile
java·后端·java ee
ChaITSimpleLove39 分钟前
.NET9 实现排序算法(MergeSortTest 和 QuickSortTest)性能测试
算法·排序算法·.net·benchmarkdotnet·datadog.trace
想躺平的咸鱼干40 分钟前
Volatile解决指令重排和单例模式
java·开发语言·单例模式·线程·并发编程
CVer儿1 小时前
svd分解求旋转平移矩阵
线性代数·算法·矩阵
雪碧聊技术1 小时前
深入解析Vue中v-model的双向绑定实现原理
前端·javascript·vue.js·v-model
快起来别睡了1 小时前
手写 Ajax 与 Promise:从底层原理到实际应用
前端