思路一

先找到最長的數組，然后循環短的數組并判斷元素是否在長數組中

public class Main {

public static void main(String[] args) throws CloneNotSupportedException {

String[] strArr1={"ee","aa","bb","cc"};

String[] strArr2={"aa","bb","cc","dd","44"};

System.out.println(Main.subContain(strArr1,strArr2));

}

public static boolean subContain(String[] strArr1, String[] strArr2) {

if (null == strArr1 || null == strArr2) {

return false;

}

boolean flag = true;

if (strArr2.length > strArr1.length) {

String[] temp;

temp = strArr1;

strArr1 = strArr2;

strArr2 = temp;

}

for (String str : strArr2) {

if (!Arrays.asList(strArr1).contains(str)) {

flag = false;

break;

}

}

return flag;

}

}

思路二

判斷兩個數組的交集長度是否等于最小的數組長度，用到了第三方的jar包

public class Main {

public static void main(String[] args) throws CloneNotSupportedException {

String[] strArr1 = {"aa", "bb", "cc"};

String[] strArr2 = {"aa", "bb", "cc", "dd", "44"};

System.out.println(Main.subContain(strArr1, strArr2));

}

public static boolean subContain(String[] strArr1, String[] strArr2) {

if (null == strArr1 || null == strArr2) {

return false;

}

Collection collection = org.apache.commons.collections.CollectionUtils.intersection(Arrays.asList(strArr1), Arrays.asList(strArr2));

return collection.size() == Math.min(strArr1.length, strArr2.length);

}

}

總結

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