java - 计算字符串中连续重复出现的字符数

Question

这是我的代码。

public static void countContinuosOccurence() {
    String first = "ABBCDDDEFGGH";
    StringBuffer result = new StringBuffer();
    int count = 1;
    for (int i = 1; i < first.length(); i++) {
        if (first.charAt(i) == (first.charAt(i - 1))) {
            count++;
        } else {
            if (count > 1) {

                result.append(String.valueOf(count) + first.charAt(i - 1));
            } else {
                result.append(first.charAt(i - 1));
            }
            count = 1;
        }
    }
    System.out.println("First String is:"+ first);
    System.out.println("Result is:" + result);
}

结果是:

First String is:ABBCDDDEFGGH
Result is:A2BC3DEF2G

缺少最后一个字符？有人可以帮我解决这个问题吗？

Answer 1

不是性能最好的，但最简单的代码:

final String in = "ABBCDDDEFGGH" + '\u0000';
final StringBuilder b = new StringBuilder();
char prev = in.charAt(0);
int rpt = 0;
for (int i = 1; i < in.length(); i++) {
  final char curr = in.charAt(i);
  if (curr == prev) rpt++;
  else {
    b.append(rpt == 0? prev : "" + (rpt + 1) + prev);
    rpt = 0; prev = curr;
  }
}
System.out.println(b);