我在 Codility 上发现这个练习时遇到了一个非常奇怪的问题,这是任务描述:
Write a function:
class Solution { public int solution(int[] A); }
that, given a non-empty zero-indexed array A of N integers, returns the minimal positive integer that does not occur in A.
For example, given:
A[0] = 1
A[1] = 3
A[2] = 6
A[3] = 4
A[4] = 1
A[5] = 2
the function should return 5.
Assume that:
N is an integer within the range [1..100,000];
each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].
Complexity:
expected worst-case time complexity is O(N);
expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).
Elements of input arrays can be modified.
还有我的代码:
class Solution {
public int solution(int[] A) {
SortedSet set = new TreeSet();
for (int i = 0; i < A.length; i++)
if (A[i] > 0)
set.add(A[i]);
Iterator it = set.iterator();
int previous = 0, element = 0;
try { previous = (int)it.next(); }
catch (NoSuchElementException e) { return 1; }
while (it.hasNext()) {
element = (int)it.next();
if (element!=(previous+1)) break;
previous=element;
}
if (previous+1 < 1) return 1;
return previous+1;
}
}
代码分析:
http://i.stack.imgur.com/IlMxP.png
我想弄清楚为什么我的代码只在那个测试中提供了错误的输出,有人能帮助我吗?
提前致谢!
最佳答案
我的解决方案得分为 100/100
// you can also use imports, for example:
// import java.util.*;
// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");
import java.util.Arrays;
class Solution {
public int solution(int[] A) {
int smallest = 1;
Arrays.sort(A);
for (int i = 0; i < A.length; i++) {
if (A[i] == smallest) {
smallest++;
}
}
return smallest;
}
}
“large_2”测试用例的时间更差,为 0.292 秒。
我会说非常好。
如果您需要解释,请联系我,以便我扩展答案:)
干杯。
关于java - 寻找缺失的整数(Codility 测试),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/29051746/