假設有一個一維數組test[]={1,2,3}和一個二維數組arr1[3][5]={{1,2,5,4,3},{3,7,1,4,2},{2,9,7,8,3}}。作為輸出所需的內容如下：test is the subset of row 0 of arr1test is the subset of row 1 of arr1test is not the subset of row 2 of arr1這是我到目前為止實現的代碼：class GFG { public static void main(String args[]) { int arr1[][] = { { 11, 1, 13, 3, 7 }, { 11, 1, 17, 7, 3 }, { 2, 5, 8, 9, 10 } }; int test[] = { 11, 3, 7, 1 }; int m = arr1.length; // rows int n = test.length; int o = arr1[0].length; // no. of elements in each row System.out.println(o); // just for testing if it works int i = 0; int j = 0; int k = 0; for (i = 0; i < n; i++) { for (j = 0; j < m && j != m; j++) { for (k = 0; k < o; k++) if (test[i] == arr1[j][k]) break; if (k == o) System.out.println("test[] is " + "not a subset of arr1 " + j + " row"); else System.out.println("test[] is " + "subset of arr1 " + j + " row"); } } }}但我得到的輸出是：我意識到這是 i 循環重復打印它，但在這種情況下我仍然沒有得到令人滿意的輸出。在這里可以做什么？或者這個問題有很多優化的實現嗎？歡迎任何建議。
2 回答
12345678_0001
TA貢獻1802條經驗 獲得超5個贊
你搞砸了循環的順序：你應該先迭代arr。這個想法是 ti 使用 flags: isSubset，當在一行中找不到某個元素時，它變為 false，contains如果當前檢查的元素在一行中，則變為 true。
可以進行一些改進（如 foreach 循環和標記中斷），但我決定保持代碼簡單。
public class GFG {
public static void main(String args[]) {
int arr[][] = { { 11, 1, 13, 3, 7 },
{ 11, 1, 17, 7, 3 },
{ 2, 5, 8, 9, 10 } };
int test[] = { 11, 3, 7, 1 };
for (int i = 0; i < arr.length; i++) {
boolean isSubset = true;
for (int j = 0; j < test.length; j++) {
boolean contains = false;
for (int k = 0; k < arr[i].length; k++) {
if (test[j] == arr[i][k]) {
contains = true;
break;
}
}
if (!contains) {
isSubset = false;
break;
}
}
if (isSubset) {
System.out.println("test[] is " + "subset of arr " + i + " row");
} else {
System.out.println("test[] is " + "not a subset of arr " + i + " row");
}
}
}
}
慕運維8079593
TA貢獻1876條經驗 獲得超5個贊
嘗試將您的解決方案分解為更小的方法，這樣您的代碼會更清晰，您可以更容易地想到發生了什么。這是一個如何完成的示例：
class GFG {
public static void main(String args[]) {
int arr1[][] = { { 11, 1, 13, 3, 7 },
{ 11, 1, 17, 7, 3 },
{ 2, 5, 8, 9, 10 } };
int test[] = { 11, 3, 7, 1 };
System.out.println(arr1[0].length); // just for testing if it works
for (int i = 0; i < arr1.length; i++) {
if (isSubset(test, arr1[i])) {
System.out.println("test[] is " + "subset of arr1 " + i + " row");
} else {
System.out.println("test[] is " + "not a subset of arr1 " + i + " row");
}
}
}
/* returns true if arr contains all elements of sub */
static boolean isSubset(int[] sub, int[] arr) {
for (int e : sub) {
if (!contains(e, arr)) return false;
}
return true;
}
/* returns true if arr contains elem */
static boolean contains(int elem, int[] arr) {
for (int e : arr) {
if (elem == e) return true;
}
return false;
}
}
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