题目：
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.
链接： http://leetcode.com/problems/multiply-strings/
4/15/2017
自己的想法太麻烦，抄别人的想法。直接用一个数组表示就清楚很多了
27ms, 91%
注意的问题：
1. 可以试试类似的题是否用长数组更好
2. 第12行k的起始位置
3. 第17-19行还需要加上result[k]之前的值
4. 第29行如何从int转为char
1 public class Solution { 2 public String multiply(String num1, String num2) { 3 if (num1.equals("0") || num2.equals("0")) return "0"; 4 else if (num1.equals("1")) return num2; 5 else if (num2.equals("1")) return num1; 6 7 int length = num1.length() + num2.length(); 8 int[] result = new int[length]; 9 int k; 10 11 for (int i = num1.length() - 1; i >= 0; i--) { 12 k = length - num1.length() + i; 13 int carry = 0; 14 int a = num1.charAt(i) - '0'; 15 for (int j = num2.length() - 1; j >= 0; j--) { 16 int b = num2.charAt(j) - '0'; 17 int current = result[k]; 18 result[k] = (a * b + carry + current) % 10; 19 carry = (a * b + carry + current) / 10; 20 k--; 21 } 22 int current = result[k]; 23 result[k] = (carry + current) % 10; 24 } 25 char[] ret; 26 if (result[0] == 0) ret = new char[length - 1]; 27 else ret = new char[length]; 28 for (int i = 0; i < ret.length; i++) { 29 ret[i] = result[0] == 0? (char)(result[i + 1] + '0'): (char)(result[i] + '0'); 30 } 31 String r = new String(ret); 32 return r; 33 } 34 }
别人的算法：
注意在i和j为下标计算时，结果数组的index为i + j和i + j + 1。而且这里都不需要carry也不需要最后的判断了，因为在内层循环体里面已经算出来了
https://discuss.leetcode.com/topic/30508/easiest-java-solution-with-graph-explanation
1 public String multiply(String num1, String num2) { 2 int m = num1.length(), n = num2.length(); 3 int[] pos = new int[m + n]; 4 5 for(int i = m - 1; i >= 0; i--) { 6 for(int j = n - 1; j >= 0; j--) { 7 int mul = (num1.charAt(i) - '0') * (num2.charAt(j) - '0'); 8 int p1 = i + j, p2 = i + j + 1; 9 int sum = mul + pos[p2]; 10 11 pos[p1] += sum / 10; 12 pos[p2] = (sum) % 10; 13 } 14 } 15 16 StringBuilder sb = new StringBuilder(); 17 for(int p : pos) if(!(sb.length() == 0 && p == 0)) sb.append(p); 18 return sb.length() == 0 ? "0" : sb.toString(); 19 }
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