我正在嘗試創(chuàng)建一個(gè)函數(shù)，將數(shù)組的所有元素與第二個(gè)數(shù)組的所有元素進(jìn)行比較，并將返回所有可能的匹配項(xiàng)，如果未找到匹配項(xiàng)則返回消息。當(dāng)我嘗試實(shí)現(xiàn)代碼時(shí)，我得到一個(gè)索引超出范圍的錯(cuò)誤。在外部 for 循環(huán)完成運(yùn)行之前，內(nèi)部 for 循環(huán)可能已達(dá)到極限。如何修改它以防止發(fā)生這種情況？Stocks[] stockList3 = new Stocks[3];stockList3[0] = new Stocks("a", 2, 1, "Buy");stockList3[1] = new Stocks("a", 3, 1, "Buy");stockList3[2] = new Stocks("a", 4, 1, "Buy");Stocks[] stockList4 = new Stocks[3];stockList4[0] = new Stocks("a", 2, 1, "Buy");stockList4[1] = new Stocks("a", 5, 1, "Buy");stockList4[2] = new Stocks("a", 4, 1, "Buy");public void matching(Stocks[] array1, Stocks[] array2) { for (int i = 0; i < array1.length; i++) { for (int j = 0; i < array2.length; j++) { if (array1[i].stockPrice == array2[j].stockPrice) { System.out.println("It's a match at $" + array1[i].stockPrice); } System.out.println("still searching..."); } System.out.println("first loop test..."); }}
2 回答
侃侃無極
TA貢獻(xiàn)2051條經(jīng)驗(yàn) 獲得超10個(gè)贊
for loops使用Setcollection 來存儲(chǔ)stockPrices數(shù)組中的一個(gè)而不是 two 怎么樣？
public static List<Stocks> matching(Stocks[] one, Stocks[] two) {
Set<Integer> stockPrices = Arrays.stream(one)
.map(stock -> stock.stockPrice)
.collect(Collectors.toSet());
return Arrays.stream(two)
.filter(stock -> stockPrices.contains(stock.stockPrice))
.collect(Collectors.toList());
}
它使用O(n)額外內(nèi)存（其中n是one.length）和O(n + m)性能時(shí)間（其中m是two.length）。
函數(shù)式編程
TA貢獻(xiàn)1807條經(jīng)驗(yàn) 獲得超9個(gè)贊
在您的 j-loop 中，您說i<array2.length的是j<array2.length
public void matching ( Stocks[] array1, Stocks[] array2){
for (int i=0; i<array1.length;i++){
for (int j=0;
j<array2.length; //this j was an i
j++){
if (array1[i].stockPrice == array2[j].stockPrice){
System.out.println("It's a match at $" + array1[i].stockPrice);
}
System.out.println("still searching...");
}
System.out.println("first loop test...");
}
}
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