深圳装修招标信息网网站优化包括哪些

直接怼代码，刚好碰到了这种需求，也是想了可久，其实想想也还是挺简单的

  public static void main(String[] args) {  // 原始字符串集合  List<String> originalList = new ArrayList<>();  originalList.add("Hello");  originalList.add("World");  originalList.add("Android");  originalList.add("Development");  // 获取字符串的最大长度  int maxLength = 0;  for (String str : originalList) {  maxLength = Math.max(maxLength, str.length());  }  // 使用循环和数组操作获取每个位置的字符集合  List<String> charListAtIndex = new ArrayList<>();  StringBuilder sb = new StringBuilder();  for (int i = 0; i < maxLength; i++) {  charListAtIndex.clear();  for (String str : originalList) {  if (i < str.length()) {  sb.setLength(0); // 清空 StringBuilder  sb.append(str.charAt(i)); // 添加当前字符  charListAtIndex.add(sb.toString()); // 将 StringBuilder 转换为 String 并添加到列表中  }  }  System.out.println("Character List at Index " + i + ": " + charListAtIndex);  }  }  

结果:

Character List at Index 0: [H, W, A, D]  
Character List at Index 1: [e, o, n, e]  
Character List at Index 2: [l, r, d, v]  
Character List at Index 3: [l, o, o, l]  
Character List at Index 4: [o, r, i, p]  
Character List at Index 5: [m, e]  
Character List at Index 6: [n, t]