1.简述:
给定两个整数,分别表示分数的分子 numerator 和分母 denominator,以 字符串形式返回小数 。
如果小数部分为循环小数,则将循环的部分括在括号内。
如果存在多个答案,只需返回 任意一个 。
对于所有给定的输入,保证 答案字符串的长度小于 104 。
示例 1:
输入:numerator = 1, denominator = 2
输出:"0.5"
示例 2:
输入:numerator = 2, denominator = 1
输出:"2"
示例 3:
输入:numerator = 4, denominator = 333
输出:"0.(012)"
2.代码实现:
class Solution {
public String fractionToDecimal(int numerator, int denominator) {
long numeratorLong = (long) numerator;
long denominatorLong = (long) denominator;
if (numeratorLong % denominatorLong == 0) {
return String.valueOf(numeratorLong / denominatorLong);
}
StringBuffer sb = new StringBuffer();
if (numeratorLong < 0 ^ denominatorLong < 0) {
sb.append('-');
}
// 整数部分
numeratorLong = Math.abs(numeratorLong);
denominatorLong = Math.abs(denominatorLong);
long integerPart = numeratorLong / denominatorLong;
sb.append(integerPart);
sb.append('.');
// 小数部分
StringBuffer fractionPart = new StringBuffer();
Map<Long, Integer> remainderIndexMap = new HashMap<Long, Integer>();
long remainder = numeratorLong % denominatorLong;
int index = 0;
while (remainder != 0 && !remainderIndexMap.containsKey(remainder)) {
remainderIndexMap.put(remainder, index);
remainder *= 10;
fractionPart.append(remainder / denominatorLong);
remainder %= denominatorLong;
index++;
}
if (remainder != 0) { // 有循环节
int insertIndex = remainderIndexMap.get(remainder);
fractionPart.insert(insertIndex, '(');
fractionPart.append(')');
}
sb.append(fractionPart.toString());
return sb.toString();
}
}
标签:yyds,denominatorLong,金典,denominator,long,append,remainder,LeetCode,numeratorLong
From: https://blog.51cto.com/u_15488507/6328004