LeetCode - Complex Number Multiplication
2017-06-27 15:08
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题目
输入两个 a+bi格式的字符串,计算它们的乘积,i^2=-1,输出结果也为a+bi的形式。如 输入”1+1i”, “1+1i” (1 + i) * (1 + i) = 1 + i^2 + 2 * i = 2i,输出0+2i
思路
从字符串中抽取出数字进行计算,计算过后再进行拼装结果字符串。代码
public String complexNumberMultiply(String a, String b) { if (a == null || b == null || a.equals("") || b.equals("")) { return ""; } // 将输入用 + 号分割开 String[] tmpa = a.split("\\+"); String[] tmpb = b.split("\\+"); // 提取出所有的数字 int numA1 = Integer.valueOf(tmpa[0]); int numA2 = Integer.valueOf(tmpa[1].replaceAll("i", "")); int numB1 = Integer.valueOf(tmpb[0]); int numB2 = Integer.valueOf(tmpb[1].replaceAll("i", "")); int res1 = -1 * numA2 * numB2 + numA1 * numB1; int res2 = numA1 * numB2 + numA2 * numB1; return res1 + "+" + res2 + "i"; }
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