經(jīng)過(guò)昨天一天，還有今天的學(xué)習(xí)，終于是把第一個(gè)程序給搞出來(lái)，雖然是借鑒他人的，但是還是收獲頗多。現(xiàn)在把

這個(gè)程序的問(wèn)題，和解決方案在下面貼出：

MultiplyStrings

Given two numbers represented as strings, return multiplication of the numbers as a string.

Note: The numbers can be arbitrarily large and are non-negative.

給定用字符串表示的兩個(gè)數(shù)字，計(jì)算其乘積：

public class MultiplyStrings {//這種算法是模擬了正常的乘積運(yùn)算解出答案。<span style="white-space:pre"> </span>//原始解法中問(wèn)號(hào)部分，我認(rèn)為是不需要的，不太理解是什么意思，<span style="white-space:pre"> </span>//我在Leecode中測(cè)試是可以通過(guò)的。public static void main(String[] args) {String a1 = "12888";String b1 = "13131555";String s = multiply(a1, b1);System.out.println("s =" + s);}public static String multiply(String num1, String num2) {StringBuilder s1 = new StringBuilder(num1).reverse();StringBuilder s2 = new StringBuilder(num2).reverse();int l1 = s1.length();int l2 = s2.length();int[] result = new int[l1 + l2 - 1];for(int i = 0; i < l1; i++){for(int j = 0; j < l2; j++){result[i+j] += (s1.charAt(i) - '0') * (s2.charAt(j) - '0');}}//System.out.println(Arrays.toString(result));StringBuilder s = new StringBuilder();int carry = 0;for(int i = 0; i < result.length; i++){int currDigit = (result[i] + carry) % 10;carry = (result[i]+carry)/10;s.insert(0, currDigit);System.out.println(s);}if(carry!=0){s.insert(0, carry);} /*???????????????????????????????????????????????????????int i = 0;while( i < s.length() && s.charAt(i) == '0'){i++;}if(i == s.length() ) s.delete(0,i-1);else s.delete(0, i); */return s.toString();} }

轉(zhuǎn)載于:https://www.cnblogs.com/sunbinbin/p/5122560.html

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