我一直在寻找这个问题的答案。到目前为止，我找到的所有东西都接近我需要的，但不完全是。我就是找不到合适的关键词组合来搜索。
我有一个定义了缺少键的数组，像这样(键是自动从其他数据生成的，可以是任何东西，尽管它们总是按从低到高的顺序排列):
$Array = array(
1 => "A",
2 => "B",
4 => "C",
7 => "D",
8 => "E",
9 => "F",
12 => "G",
15 => "H",
16 => "I"
);
我需要创建另一个数组来检测每组按顺序编号的键，如下所示:
Array
(
[0] => Array
(
[0] => 1
[1] => 2
)
[1] => Array
(
[0] => 7
[1] => 8
[2] => 9
)
[2] => Array
(
[0] => 15
[1] => 16
)
)
键"4"answers"12"将被丢弃，因为"5"answers"13"分别不存在。
我怎样才能做到这一点?
我已经尝试了我能想到的for、foreach和while循环迭代的所有组合。我总是得到完全混乱的，过长的数组，看起来像
[ . . . ]
[16] => Array
(
[0] => 7
[1] => 8
[2] => 9
)
[19] => Array
(
[0] => 4
)
[21] => Array
(
[0] => 7
[1] => 8
[2] => 9
)
[27] => Array
(
[0] => 8
[1] => 9
)
[30] => Array
(
[0] => 9
)
[33] => Array
(
[0] => 8
[1] => 9
)
[ . . . ]
我在想一个更好的方法，但这是有效的:
$i = 0;
foreach($Array as $k => $v) {
if(isset($Array[$k+1])) {
$result[$i][] = $k;
$result[$i][] = $k+1;
$result[$i] = array_unique($result[$i]);
} else {
$i++;
}
}
粘合在一起，但需要进一步测试:
$Array = array(
1 => "A",
2 => "B",
4 => "C",
7 => "D",
8 => "E",
9 => "F",
12 => "G",
15 => "H",
16 => "I"
);
$ret = array();
$cur = array();
print_r($Array);
$prevKey = NULL;
foreach($Array as $k => $v)
{
if (!isset($prevKey) || $k - $prevKey == 1)
{
$cur[] = $k;
} else {
$ret[] = $cur;
$cur = array();
$cur[] = $k;
}
$prevKey = $k;
}
if (count($cur) > 0)
$ret[] = $cur;
print_r($ret);
Array
(
[1] => A
[2] => B
[4] => C
[7] => D
[8] => E
[9] => F
[12] => G
[15] => H
[16] => I
)
Array
(
[0] => Array
(
[0] => 1
[1] => 2
)
[1] => Array
(
[0] => 4
)
[2] => Array
(
[0] => 7
[1] => 8
[2] => 9
)
[3] => Array
(
[0] => 12
)
[4] => Array
(
[0] => 15
[1] => 16
)
)
这是特定的有一个现有的PHP函数。这应该按照您指定的方式生成结果…
$Array = array(
1 => "A",
2 => "B",
4 => "C",
7 => "D",
8 => "E",
9 => "F",
12 => "G",
15 => "H",
16 => "I"
);
$lastNum = -1;
$sequences = array();
$curSequence = array();
foreach($Array as $arrayIndex => $arrayValue){
if($arrayIndex == $lastNum + 1 || $lastNum === -1){
//Is sequential...
$curSequence[] = $arrayIndex;
}else{
//Non-sequential...
if(count($curSequence) > 1){
//Add current sequence
$sequences[] = $curSequence;
}
//Start a new sequence
$curSequence = array($arrayIndex);
}
$lastNum = $arrayIndex;
}
//At the end of the loop, store current sequence
if(count($curSequence) > 1){
$sequences[] = $curSequence;
}
echo "<pre>"; print_r($sequences); echo "</pre>";
Array
(
[0] => Array
(
[0] => 1
[1] => 2
)
[1] => Array
(
[0] => 7
[1] => 8
[2] => 9
)
[2] => Array
(
[0] => 15
[1] => 16
)
)
问题标记为PHP但如果有人感兴趣，这里有一个Java解决方案
import java.util.*;
public class FindSequential {
public static void main(String args[]) {
Map<Integer, String> inputMap = new LinkedHashMap<Integer, String>();
inputMap.put(1,"A");
inputMap.put(2,"B");
inputMap.put(4,"C");
inputMap.put(5,"D");
inputMap.put(8,"E");
inputMap.put(9,"F");
inputMap.put(12,"G");
inputMap.put(15,"H");
inputMap.put(16,"I");
inputMap.put(18,"J");
inputMap.put(20,"K");
inputMap.put(21,"L");
Map<Integer, List<Integer>> outputList = new HashMap<Integer, List<Integer>>();
int previousKey = -1;
int outputListIndex = 0;
List<Integer> tempList = null;
for (int key: inputMap.keySet()) {
if(previousKey == -1) {
// first element
previousKey = key;
tempList = new ArrayList<Integer>();
continue;
}
int diff = key - previousKey;
if(diff == 1) {
// continuous sequence
if(!tempList.contains(previousKey)) {
tempList.add(previousKey);
}
if(!tempList.contains(key)) {
tempList.add(key);
}
outputList.put(outputListIndex,tempList);
} else {
// not a sequence
if(tempList.size() > 0) {
outputListIndex++;
}
tempList = new ArrayList<Integer>();
}
previousKey = key;
}
// you can print the outputList here
}
}