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Leetcode 166. Fraction to Recurring Decimal

發布時間:2025/5/22 编程问答 26 豆豆
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這個是我比較原始的解法。在這里有個trick就是把int擴展為long。要考慮負數情況,以及負數最小值的情況。

public class Solution {public String fractionToDecimal(int numerator, int denominator) {long lnumerator = numerator;long ldenominator = denominator;if (lnumerator == 0) {return "0";}boolean isPositive = true;if ((lnumerator >0 && ldenominator<0) || (lnumerator <0 && ldenominator>0)) {isPositive = false;}if (lnumerator < 0) {lnumerator = -lnumerator;}if (ldenominator < 0) {ldenominator = -ldenominator;}String left = "";if (lnumerator >= ldenominator && lnumerator % ldenominator == 0) {left = String.valueOf(lnumerator / ldenominator);if (!isPositive) {left = "-" + left;}return left;}if (denominator != 0) {long zhengshu = lnumerator / ldenominator;lnumerator = lnumerator % ldenominator;List<Long> yushus = new ArrayList<Long>();while (true) {long shang = lnumerator / ldenominator;long yushu = lnumerator % ldenominator;left = left + String.valueOf(shang);lnumerator = yushu * 10;if (shang == 0) {yushus.add(yushu);continue;} else if (yushu == 0) {left = String.valueOf(zhengshu) + "." + left.substring(1);if (!isPositive) {left = "-" + left;}return left;} else {for (int i = 0; i < yushus.size(); i++) {if (yushus.get(i) == yushu) {left = String.valueOf(zhengshu) + "." + left.substring(1, i+1) + "(" + left.substring(i+1) + ")";if (!isPositive) {left = "-" + left;}return left;}}}yushus.add(yushu);}}return left;} }

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