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分享诚征大神求解codility上两个测试题，智力和能力大考验！（我是被打击惨了，看看大神们如何吧）
不说了，有机会在codility上做了两个测试题，用php解。我写了两个程序，当时感觉还行，但出来结果一看，200分就得了50分。诚征大神指出我的程序中的错误，以及大神们的solution.
啥都不说了，先上题目：
1. You are given a non-empty zero-indexed array A consisting of N integers. Each element of array A can be −1, 0 or 1.
A pair of integers (P, Q), such that 0 ≤ P ≤ Q < N, is called a slice of array A. The sum of a slice (P, Q) is the total of A[P] + A[P + 1] + ... + A[Q].
You are given integer S. The goal is to find the largest slice whose sum equals S.
For example, consider integer S = 2 and array A such that:
A[0] = 1
A[1] = 0
A[2] = -1
A[3] = 1
A[4] = 1
A[5] = -1
A[6] = -1
There are exactly two slices (0, 4) and (3, 4) whose sum equals 2. The largest such slice is (0, 4) and its length equals 5.
Write a function:
function solution($A, $S);
that, given a non-empty zero-indexed array A consisting of N integers and an integer S, returns the length of the largest slice whose sum equals S.
The function should return −1 if there is no such slice.
For example, given S = 2 and:
A[0] = 1
A[1] = 0
A[2] = -1
A[3] = 1
A[4] = 1
A[5] = -1
A[6] = -1
the function should return 5, as explained above.
Assume that:
N and S are integers within the range [1..100,000];
each element of array A is an integer within the range [−1..1].
Complexity:
expected worst-case time complexity is O(N);
expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).
Elements of input arrays can be modified.
2.A non-empty zero-indexed array A consisting of N integers is given.
You can perform a single swap operation in array A. This operation takes two indices I and J, such that 0 ≤ I ≤ J < N, and exchanges the values of A[I] and A[J].
The goal is to check whether array A can be sorted into non-decreasing order by performing only one swap operation.
For example, consider array A such that:
A[0] = 1
A[1] = 3
A[2] = 5
A[3] = 3
A[4] = 7
After exchanging the values A[2] and A[3] we obtain an array [1, 3, 3, 5, 7], which is sorted in non-decreasing order.
Write a function:
function solution($A);
that, given a non-empty zero-indexed array A consisting of N integers, returns true if the array can be sorted into non-decreasing order by one swap operation or false otherwise.
For example, given:
A[0] = 1
A[1] = 3
A[2] = 5
A[3] = 3
A[4] = 7
the function should return true, as explained above.
On the other hand, for the following array:
A[0] = 1
A[1] = 3
A[2] = 5
A[3] = 3
A[4] = 4
the function should return false, as there is no single swap operation that sorts the array.
Assume that:
N is an integer within the range [1..100,000];
each element of array A is an integer within the range [1..1,000,000,000].
Complexity:
expected worst-case time complexity is O(N*log(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.
我的程序：
针对第一题：
针对第二题：
我是被打击惨了，最后200分中只得了50分。大神们，你们能得到多少分？
啥都不说了，先上题目：
1. You are given a non-empty zero-indexed array A consisting of N integers. Each element of array A can be −1, 0 or 1.
A pair of integers (P, Q), such that 0 ≤ P ≤ Q < N, is called a slice of array A. The sum of a slice (P, Q) is the total of A[P] + A[P + 1] + ... + A[Q].
You are given integer S. The goal is to find the largest slice whose sum equals S.
For example, consider integer S = 2 and array A such that:
A[0] = 1
A[1] = 0
A[2] = -1
A[3] = 1
A[4] = 1
A[5] = -1
A[6] = -1
There are exactly two slices (0, 4) and (3, 4) whose sum equals 2. The largest such slice is (0, 4) and its length equals 5.
Write a function:
function solution($A, $S);
that, given a non-empty zero-indexed array A consisting of N integers and an integer S, returns the length of the largest slice whose sum equals S.
The function should return −1 if there is no such slice.
For example, given S = 2 and:
A[0] = 1
A[1] = 0
A[2] = -1
A[3] = 1
A[4] = 1
A[5] = -1
A[6] = -1
the function should return 5, as explained above.
Assume that:
N and S are integers within the range [1..100,000];
each element of array A is an integer within the range [−1..1].
Complexity:
expected worst-case time complexity is O(N);
expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).
Elements of input arrays can be modified.
2.A non-empty zero-indexed array A consisting of N integers is given.
You can perform a single swap operation in array A. This operation takes two indices I and J, such that 0 ≤ I ≤ J < N, and exchanges the values of A[I] and A[J].
The goal is to check whether array A can be sorted into non-decreasing order by performing only one swap operation.
For example, consider array A such that:
A[0] = 1
A[1] = 3
A[2] = 5
A[3] = 3
A[4] = 7
After exchanging the values A[2] and A[3] we obtain an array [1, 3, 3, 5, 7], which is sorted in non-decreasing order.
Write a function:
function solution($A);
that, given a non-empty zero-indexed array A consisting of N integers, returns true if the array can be sorted into non-decreasing order by one swap operation or false otherwise.
For example, given:
A[0] = 1
A[1] = 3
A[2] = 5
A[3] = 3
A[4] = 7
the function should return true, as explained above.
On the other hand, for the following array:
A[0] = 1
A[1] = 3
A[2] = 5
A[3] = 3
A[4] = 4
the function should return false, as there is no single swap operation that sorts the array.
Assume that:
N is an integer within the range [1..100,000];
each element of array A is an integer within the range [1..1,000,000,000].
Complexity:
expected worst-case time complexity is O(N*log(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.
我的程序：
针对第一题：
function solution($A, $S) {
// write your code in PHP5.3
$len = count($A);
$slice = -1;
for ($begin=0; $begin<$len;$begin++)
{
$total = 0;
for ($end = $begin; $end < $len; $end ++)
{
$total = $total + $A[$end];
if ($total == $S)
{
$temp = $end - $begin +1;
if ($temp > $slice)
{
$slice = $temp;
}
}
}
}
return $slice;
}
针对第二题：
function solution($A) {
// write your code in PHP5.3
$arr = array();
$len = count($A);
$check = false;
for ($i=0;$i<$len-1;$i++)
{
$arr = $A;
for ($j=$i+1;$j<$len;$j++)
{
$temp = $arr[$j];
$arr[$j] = $arr[$i];
$arr[$i] = $temp;
$no_decrease = true;
for ($k=1;$k< $len;$k++)
{
if ($arr[$k]<$arr[$k-1])
{
$no_decrease = false;
}
}
if ($no_decrease == true)
{
$check = true;
}
}
}
return $check;
}
我是被打击惨了，最后200分中只得了50分。大神们，你们能得到多少分？
...全文
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xuefei_wen 2015-10-05
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我一看到O(NlogN)第一反应就是排序 我觉得这道题可以用复制数组然后进行排序 比较排序后的数组和排序前的 如果只有两个位置的数不一样，或者全都一样且数组中有重复元素，就返回true,否则返回false
lhw7791086 2014-10-10
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xx87343133 2014-10-08
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试试[6,3,4,2,7],这是可以的，交换6,2，但程序结果是false
[/quote]
nettman 2014-10-05
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xuzuning 2014-10-03
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再看第一题
同样计算的结果并没有错误，只是算法没有符合：时间复杂度为 O(N) 的要求
你使用了双重循环，所以时间复杂度为 O(N*N)
可以用闭包来去掉一重循环
var_dump(solution([1,0,-1,1,1,-1,-1], 2));
function solution($A, $S) {
$len = count($A);
$slice = -1;
for ($begin=0; $begin<$len; $begin++) {
$total = 0;
array_walk($A, function($v, $end, $S) use ($begin, &$slice, &$total) {
if($end < $begin) return;
$total += $v;
if($total == $S) {
$temp = $end - $begin + 1;
if ($temp > $slice) $slice = $temp;
}
}, $S);
}
return $slice;
}xuzuning 2014-10-03
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先看第二题
原代码计算结果是正确的，而#1的补充是错误的！
如果将 $arr = $A; 放在第二个 for 的后面那么将出现一系列的使用未定义变量的警告
解答被扣分的原因在于你的答案不符合：时间复杂度为 O(N*log(N)) 的要求
可以写作这样
var_dump(solution([1,3,5,3,7]));
var_dump(solution([1,3,5,3,4]));
function solution($A) {
$arr = array();
$len = count($A);
$check = 0;
for($i=0;$i<$len-1;$i++) {
if($A[$i+1] < $A[$i]) {
$temp = $A[$i+1];
$A[$i+1] = $A[$i];
$A[$i] = $temp;
if($i > 0 && $A[$i-1] <= $A[$i]) $check++;
}
}
return $check == 1;
}bool(true)
bool(false)
歪着看世界 2014-10-03
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毫无疑问，xuzuning 版主是我心目中努力的一个目标，程序算法犀利无比，再一次感受到啊。
啥也别说，肯定100分啊。
对了，我查了一些资料，在引用codility题目时一般都隐去题目，因为涉及到版权问题。所以，xuzuning版主，是不是把题目具体文字隐去？或者其他方式处理？ 只是建议啊。
xuzuning 2014-10-03
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修改后如下
var_dump(solution([1,3,5,3,7]));
var_dump(solution([1,3,5,3,4]));
var_dump(solution([1,6,3,4,3,7]));
var_dump(solution([1,6,5,3,3,4,7]));
function solution($A) {
$arr = array();
$len = count($A);
$check = 0;
for($i=0;$i<$len-1;$i++) {
if($A[$i+1] < $A[$i]) { //如果降序
$flag = 0;
for($j=$i+1; $j<$len-1; $j++) { //向后找到符合升序的位置
if($A[$i] < $A[$j+1]) {
$flag = 1;
break;
}
}
$temp = $A[$j]; //交换
$A[$j] = $A[$i];
$A[$i] = $temp;
if($i > 0 && $A[$i-1] <= $A[$i]) $check++; //交换成功
else return false; //交换失败
if(! $flag) $i--; // $flag = 0 表示前面的查找是自然结束的，有待商榷
}
}
return $check == 1;
}xuzuning 2014-10-03
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if($i > 0 && $A[$i-1] <= $A[$i]) $check++; //交换成功，记录
else return false; //交换失败，返回。 原先漏掉了
if($A[$i] < $A[$j+1]) {
$flag = 1;
break;
}
的意思是找到第一个大于 $A[$i] 的位置
要不要等号无所谓
歪着看世界 2014-10-03
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仔细测试了一下你的代码。短时间还没有办法理解版主的思路，不过我多测试了几个例子，大部分正确，但依然有问题。
比如这个例子，var_dump(solution([1,6,5,3,3,4,7]));
目测应该无法一次调换成功，但程序返回是正确的。
因为暂时还没有理解版主大神的思路，所以还不好评论，但我一直担心non-decreasing 非降，其中包括等号=的情况，比如下面
if($A[$i] < $A[$j+1]) {
$flag = 1;
break;
}
其中，$A[$i] < $A[$j+1]是否需要等号呢？
我提供的测试例子是不是应该返回false呢？
xuzuning 2014-10-03
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时间复杂度的简单判断
for() {
code
}
==> O(N)
for() {
code
for() {
code
}
}
==> O(N*log(N))
for() {
for() {
code
}
}
==> O(N*N)
xuzuning 2014-10-03
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对于第二题，我只是根据题目的示例而为之，的确没有考虑的 交换的位置不相邻的情况
所以也在纳闷为何是 O(N*log(N)) 呢？O(N)) 不就可以了吗
改写一下
var_dump(solution([1,3,5,3,7]));
var_dump(solution([1,3,5,3,4]));
var_dump(solution([1,6,3,4,3,7]));
function solution($A) {
$arr = array();
$len = count($A);
$check = 0;
for($i=0;$i<$len-1;$i++) {
if($A[$i+1] < $A[$i]) {
$flag = 0;
for($j=$i+1; $j<$len-1; $j++) {
if($A[$i] < $A[$j+1]) {
$flag = 1;
break;
}
}
$temp = $A[$j];
$A[$j] = $A[$i];
$A[$i] = $temp;
if($i > 0 && $A[$i-1] <= $A[$i]) $check++;
if(! $flag) $i--;
}
}
return $check == 1;
}bool(true)
bool(false)
bool(true)
歪着看世界 2014-10-03
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就没有大神能来解决么？ 尤其第二道题，还是蛮有难度的。
歪着看世界 2014-10-03
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版主第一题的解法，我验证过，确实正确，感觉用这个巧妙的函数去掉一次循环。
我查了array_walk的介绍，还不能理解版主精妙的思路。版主能否提点两句？
还有，第二个题目，调整后的代码如下，供版主以及其他大神测试用
$arr = array();
$len = count($A);
$check = false;
for ($i=0;$i<$len-1;$i++)
{
for ($j=$i+1;$j<$len;$j++)
{
$arr = $A;
$temp = $arr[$j];
$arr[$j] = $arr[$i];
$arr[$i] = $temp;
$no_decrease = true;
for ($k=1;$k< $len;$k++)
{
if ($arr[$k]<$arr[$k-1])
{
$no_decrease = false;
}
}
if ($no_decrease == true)
{
$check = true;
}
}
}
return $check;
歪着看世界 2014-10-03
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感谢版主大驾光临啊。
不过对于第二题，我有不同看法：
1.你的思路是：检查相邻两个大小，若右边的小于左边的数值，则调换；且这种调换满足i左边的大小关系，则check++.
事实上，我在做题时考虑过这个思路，不过考虑这个例子：1,6,3,4,3,7.只要把6和第二个3调换一下，则可以满足非降数列，最后应该是true。但是，按照你的算法，这个题目返回false。
关键点是：数字的调换可以是很巧妙的一次调换。
2.第二题中，$arr = $A;应该放在第二个for后面，我又验证了一遍，不会出现未定义对象等错误。结果是正确的。
我的思路是每次把$A赋值给$arr，并且采用穷举法把任意两个不同数字调换一次，判断调换后的结果是不是非降数列。是的话，返回true，否的话，不改变check的false值。只要穷举法中，任意一次可以做到，check的值就变为true.
这个思路是正确的，但就是时间复杂度不满足，扣分了。
版主，你看呢？
天涯过客009 2014-10-02
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第天回贴即可获得１０分可用分
sdgdh43 2014-10-02
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不识英语。。 0分
歪着看世界 2014-10-02
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尼玛，仔细一看，第二题我的程序就有个明显错误，$arr = $A;应该放在第二个 for 的后面。
晕死。
