public int findLast (int[] x, int y) {

//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 (int i=x.length-1; i > 0; i--)

{

if (x[i] == y)

{

return i;

}

}

return -1;

}

// test: x=[2, 3, 5]; y = 2

// Expected = 0

 

1.不会遍历第一个元素，i应该大于等于0

2.DO NOT EXECUTE THE FAULT  test:x=null,y=3   Expected=NullPointerException

3.EXECUTE THE FAULT BUT NOT RESULT IN AN ERROR STATE 

test:xx=[1,2,3],y=3  Expected=2

4.RESULT IN AN ERROR BUT NOT A FAILURE  test=[1,2,3],y=4  Expected=-1

 

public static int lastZero (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 (int i = 0; i < x.length; i++)

{

if (x[i] == 0)

{

return i;

}

} return -1;

}

// test: x=[0, 1, 0]

// Expected = 2

 

1.遍历顺序写反了，for循环应改为for(int i=x.length-1;i>=0;i--)

2.test: x=null  Expected=NullPointerException

3.test: x=[1]   Expected=-1

4.test: x=[1,2,0]  Expected=2