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1, findLast()
(1) The fault is that the state of the loop should be:
for(int i = x.length-1;i >= 0;i--)
Because if the expected index is in x[0], when i=0 the loop will not be execute because it doesn’t fit the state that I > 0.
(2) We must make sure it will not execute the i>0, so let x be null,
Test case: x=null y=2
Expected: NullPointerException
Actual: NullPointerException
(3) execute the fault but not cause error:
Test case: x = [1,2,5] y = 2
Expected: 1
Actual: 1
(4) result error but not a failure:
Test case: x = [1,2,3] y = 5
Expected: -1
Actual: -1
To cause the error, we should let it execute the state i = 0, but I = 0 will cause the return -1, because it will be push out of the loop, at the same time ,there is no expected number in the array, so the actual result also is -1.
2,lastZero()
(1) The question says that finding the last zero, so we should start with the last index in the array, so for(int i = x.length – 1;i>=0;i--)
(2) Whatever the case is ,the int i=0 will be execute at first.
(3) Test case: x = [2,1,4]
Expected: i = -1
Actual: i = -1
(4)Test case: x = [0,1,2]
Expected: i = 0
Actual: I = 0
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原文地址：http://www.cnblogs.com/mycode2016/p/5257529.html
