大整數(shù)相乘遞歸 分治思想

1、package com.liheng.algorithm;import java.util.Scanner;import java.util.regex.Matcher;import java.util.regex.Pattern/* * 大整數(shù)乘法 * * */public class BigIntMultiply /規(guī)模只要在這個(gè)范圍內(nèi)就可以直接計(jì)算了 private final static int SIZE = 4; / 此方法要保證入?yún)en為X、Y的長(zhǎng)度最大值 private static String bigIntMultiply(String X, String Y, int

2、len) / 最終返回結(jié)果 String SA_B,SD_C; String str = "" int flag=0; / 補(bǔ)齊X、Y，使之長(zhǎng)度相同 X = formatNumber(X, len); Y = formatNumber(Y, len); / 少于4位數(shù)，可直接計(jì)算 if (len <= SIZE) return "" + (Integer.parseInt(X) * Integer.parseInt(Y); / 將X、Y分別對(duì)半分成兩部分 int len1 = len / 2; int len2 = len - len1; Stri

3、ng A = X.substring(0, len1); String B = X.substring(len1); String C = Y.substring(0, len1); String D = Y.substring(len1); / 兩字符串長(zhǎng)度要相同 int lenM = Math.max(A.length(), B.length(); String A1 = formatNumber(A, lenM); String B1 = formatNumber(B, lenM); int a=Integer.parseInt(A1.substring(0, 1); int b= In

4、teger.parseInt(B1.substring(0, 1); if(a>b) SA_B=dudcting(A1,B1); else SA_B = dudcting(B1,A1); flag+; lenM = Math.max(A.length(), B.length(); String C1 = formatNumber(C, lenM); String D1 = formatNumber(D, lenM); int c=Integer.parseInt(C1.substring(0, 1); int d= Integer.parseInt(D1.substring(0, 1);

5、 if(d>c) SD_C=dudcting(D1,C1); else SD_C = dudcting(C1,D1); flag+; lenM=Math.max(SA_B.length(), SD_C.length(); String ABDC="" if(flag!=1) ABDC = bigIntMultiply(SA_B, SD_C, lenM); else ABDC = bigIntMultiply(SA_B, SD_C, lenM); String AC = bigIntMultiply(A, C, len1); String BD = bigIntMult

6、iply(B, D, len2); / 處理BD，得到原位及進(jìn)位 String sBD = dealString(BD, len2); / 處理AD+BC的和 String ADBC = addition(AC, BD); String ABCD="" lenM = Math.max(ABDC.length(), ADBC.length(); if(flag=1) String ABDC1 = formatNumber(ABDC, lenM); String ADBC1 = formatNumber(ADBC, lenM); int a1=Integer.parseInt(

7、ABDC1.substring(0, 1); int b1= Integer.parseInt(ADBC1.substring(0, 1); ABCD = dudcting(ADBC1,ABDC1); else ABCD = addition(ABDC,ADBC); flag=0; / 加上BD的進(jìn)位 if (!"0".equals(sBD1) ABCD = addition(ABCD, sBD1); / 得到ADBC的進(jìn)位 lenM = Math.max(len1, len2); String sABCD = dealString(ABCD, lenM); / AC加上A

8、DBC的進(jìn)位 AC = addition(AC, sABCD1); / 最終結(jié)果 str = AC + sABCD0 + sBD0; return str; / 兩個(gè)數(shù)字串按位加 private static String addition(String ad, String bc) / 返回的結(jié)果 String str = "" / 兩字符串長(zhǎng)度要相同 int lenM = Math.max(ad.length(), bc.length(); ad = formatNumber(ad, lenM); bc = formatNumber(bc, lenM); / 按位加，進(jìn)

9、位存儲(chǔ)在temp中 int flag = 0; / 從后往前按位求和 for (int i = lenM - 1; i >= 0; i-) int t = flag + Integer.parseInt(ad.substring(i, i + 1) + Integer.parseInt(bc.substring(i, i + 1); / 如果結(jié)果超過9，則進(jìn)位當(dāng)前位只保留個(gè)位數(shù) if (t > 9) flag = 1; t = t - 10; else flag = 0; / 拼接結(jié)果字符串 str = "" + t + str; if (flag != 0)

10、str = "" + flag + str; return str; / 兩個(gè)數(shù)字串按位減 private static String dudcting(String ad, String bc) / 返回的結(jié)果 String str = "" / 按位加，進(jìn)位存儲(chǔ)在temp中 int flag = 0; / 從后往前按位求和 for (int i = ad.length() - 1; i >= 0; i-) int t =flag+Integer.parseInt(ad.substring(i, i + 1)-Integer.parseInt(b

11、c.substring(i, i + 1); / 如果結(jié)果超過9，則進(jìn)位當(dāng)前位只保留個(gè)位數(shù) if (t <0) flag = -1; t = t+10; else flag = 0; / 拼接結(jié)果字符串 str = "" + t + str; if (flag != 0) str = "" + flag + str; return str; / 處理數(shù)字串，分離出進(jìn)位； / String數(shù)組第一個(gè)為原位數(shù)字，第二個(gè)為進(jìn)位 private static String dealString(String ac, int len1) String str

12、 = ac, "0" if (len1 < ac.length() int t = ac.length() - len1; str0 = ac.substring(t); str1 = ac.substring(0, t); else / 要保證結(jié)果的length與入?yún)⒌膌en一致，少于則高位補(bǔ)0 String result = str0; for (int i = result.length(); i < len1; i+) result = "0" + result; str0 = result; return str; / 乘數(shù)、被乘數(shù)

13、位數(shù)對(duì)齊 private static String formatNumber(String x, int len) while (len > x.length() x = "0" + x; return x; /測(cè)試樁 public static void main(String args) / 正則表達(dá)式：不以0開頭的數(shù)字串 String pat = "1-9d*$" Pattern p = Ppile(pat); / 獲得乘數(shù)A System.out.println("請(qǐng)輸入乘數(shù)A（不以0開頭的正整數(shù)）："); Scanner sc = new Scanner(System.in); String A = sc.nextLine(); Matcher m = p.matcher(A); if (!m.matches() System.out.println("數(shù)字不合法！"); return; / 獲得乘數(shù)B System.out.println("請(qǐng)輸入乘數(shù)B（不以0開頭的正整數(shù)）："); sc = new Scanner(System.in); String B