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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
回答：
1.检测不到x[0]
2.令x = NULL
3.x = [2,3]; y = 3
4.x = [2,3]; y = 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
回答：
1.要求返回最后一个零的位置，代码返回第一个零的位置
2.x = NULL;
3.x = [1,0];
4.x = [0,1];
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原文地址：http://www.cnblogs.com/c337134154/p/5254253.html
