軟體測試homework2

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一.

  

public intfindLast(int[] x, inty) {
//Effects: If x==null throw
NullPointerException
// else return the index of the last element
// in x that equals y.
// If no such element exists, return -1
for (inti=x.length-1; i> 0; i--)
{
if (x[i] == y)
{
return i;
}
}
return -1;
}
// test: x=[2, 3, 5]; y = 2
// Expected = 0

 

1. i>0 改為 i>=0

2.x=[3,4,5];y=5

3.x=[3,4,5];y=6

4.x=[3,4,5];y=3

 

public static intlastZero(int[] x) {
//Effects: if x==null throw
NullPointerException
// else return the index of the LAST 0 in x.
// Return -1 if 0 does not occur in x
for (inti= 0; i< x.length; i++)
{
if (x[i] == 0)
{
return i;
}
} return -1;
}
// test: x=[0, 1, 0]
// Expected = 2

1.for (inti= 0; i< x.length; i++) 改為 for (int i=x.length-1; i>= 0; i--)

2.x=[0,4,5];y=0

 

軟體測試homework2