寫在前面

該方法目前只能處理Integer能夠裝得下的值，已經處理了【零...】的情況

有實現思路，僅供參考，發現問題可調優的地方多謝指正

上菜!!!

package com.xxx.util;import javax.validation.constraints.NotNull; import java.util.regex.Matcher; import java.util.regex.Pattern;public class StringUtil {private static final String[] CHINESE_NUMBER = {"零", "一", "二", "三", "四", "五", "六", "七", "八", "九"};private static final String[] UNIT = {"", "十", "百", "千"};private static final String[] MORE_THAN_TEN_THOUSAND = {"", "萬", "億"};private static final Pattern zeroPattern = Pattern.compile("[\u96f6]+");public static String convertArabNumToChineseNum(@NotNull Integer arabNum) {String[] split = arabNum.toString().split("");/*** step 1:第一次處理，只需要將數字轉為漢字即可*/StringBuffer dealNum = new StringBuffer();for (int i = 0; i < split.length; i++) {dealNum.append(CHINESE_NUMBER[Integer.parseInt(split[i])]);}/*** step 2:將第一次處理后的數字保存下來，并清空stringBuffer以供后面使用，也可以重新聲明一個*/String firstDealNum = dealNum.toString();dealNum.setLength(0);/*** step 3:將數字從后往前面取，每四位取一次，然后轉換成千百十的數字，比如1234就轉換為一千二百三十四* step 4:將數字拼接起來，順帶加上萬億等單位 count為是否已經達到萬或者億的數量級*/int count = 0;for (int i = firstDealNum.length(); i > 0; i -= 4) {String substring = firstDealNum.substring(i - 4 < 0 ? 0 : i - 4, i);String[] numbers = substring.split("");StringBuffer sb = new StringBuffer();for (int j = 0; j < numbers.length; j++) {String num = numbers[j];String unit = UNIT[numbers.length - 1 - j];if (CHINESE_NUMBER[0].equals(num)) {sb.append(num);}else{sb.append(num).append(unit);}}dealNum.insert(0, MORE_THAN_TEN_THOUSAND[count]);dealNum.insert(0, sb.toString());count++;}/*** step 5:個位是零的話給他干掉* step 6:將數字中的【零】【零...】替換為【零】*/if (dealNum.length() - 1 == dealNum.lastIndexOf("零")) {dealNum.deleteCharAt(dealNum.length() - 1);}Matcher matcher = zeroPattern.matcher(dealNum.toString());return matcher.replaceAll("零");}}

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