如何在没有java utils的情况下比较两个字符串数组
检查数组arr1是否包含与java中相同顺序的arr2相同的元素.
Check to see if the array arr1 contain the same elements as arr2 in the same order in java.
例如:
isTheSame({"1", "2", "3"}, {"1", "2", "3"}) → true
isTheSame({"1", "2", "3"}, {"2", "1", "1"}) → false
isTheSame({"1", "2", "3"}, {"3", "1", "2"}) → false
目前为止
public boolean isTheSame(String[] arr1, String[] arr2)
{
if (arr1.length == arr2.length)
{
for (int i = 0; i < arr1.length; i++)
{
if (arr1[i] == arr2[i])
{
return true;
}
}
}
return false;
}
这样做的问题是它只比较两个数组的第一个元素.
The problem with this is that it only compares the first element of the two arrays.
推荐答案
您一直在迭代,直到找到匹配项.你应该寻找一个不匹配的字符串,你应该使用 equals
而不是 ==
You are iterating until you find a match. You should instead be looking for a String which doesn't match and you should be using equals
not ==
// same as Arrays.equals()
public boolean isTheSame(String[] arr1, String[] arr2) {
if (arr1.length != arr2.length) return false;
for (int i = 0; i < arr1.length; i++)
if (!arr1[i].equals(arr2[i]))
return false;
return true;
}
仅供参考,这就是 Arrays.equals 在处理 null
值时所做的事情.
FYI This is what Arrays.equals does as it handle null
values as well.
public static boolean equals(Object[] a, Object[] a2) {
if (a==a2)
return true;
if (a==null || a2==null)
return false;
int length = a.length;
if (a2.length != length)
return false;
for (int i=0; i<length; i++) {
Object o1 = a[i];
Object o2 = a2[i];
if (!(o1==null ? o2==null : o1.equals(o2)))
return false;
}
return true;
}
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