java 表情符过滤

担心循环判断字符会影响效率，实际上使用 速度挺好。

以下为转载内容

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import org.apache.commons.lang3.StringUtils;
 
public class EmojiFilter {
 
	/**
	 * 检测是否有emoji字符
	 * 
	 * @param source
	 * @return 一旦含有就抛出
	 */
	public static boolean containsEmoji(String source) {
		if (StringUtils.isBlank(source)) {
			return false;
		}
 
		int len = source.length();
 
		for (int i = 0; i < len; i++) {
			char codePoint = source.charAt(i);
 
			if (isEmojiCharacter(codePoint)) {
				// do nothing，判断到了这里表明，确认有表情字符
				return true;
			}
		}
 
		return false;
	}
 
	private static boolean isEmojiCharacter(char codePoint) {
		return (codePoint == 0x0) || (codePoint == 0x9) || (codePoint == 0xA)
				|| (codePoint == 0xD)
				|| ((codePoint >= 0x20) && (codePoint <= 0xD7FF))
				|| ((codePoint >= 0xE000) && (codePoint <= 0xFFFD))
				|| ((codePoint >= 0x10000) && (codePoint <= 0x10FFFF));
	}
 
	/**
	 * 过滤emoji 或者 其他非文字类型的字符
	 * 
	 * @param source
	 * @return
	 */
	public static String filterEmoji(String source) {
 
		if (!containsEmoji(source)) {
			return source;// 如果不包含，直接返回
		}
		// 到这里铁定包含
		StringBuilder buf = null;
 
		int len = source.length();
 
		for (int i = 0; i < len; i++) {
			char codePoint = source.charAt(i);
 
			if (isEmojiCharacter(codePoint)) {
				if (buf == null) {
					buf = new StringBuilder(source.length());
				}
 
				buf.append(codePoint);
			} else {
			}
		}
 
		if (buf == null) {
			return source;// 如果没有找到 emoji表情，则返回源字符串
		} else {
			if (buf.length() == len) {// 这里的意义在于尽可能少的toString，因为会重新生成字符串
				buf = null;
				return source;
			} else {
				return buf.toString();
			}
		}
 
	}
 
}