确定列表编号是否是连续的 [英] Determining if a list numbers are sequential
问题描述
我正在使用Java.我有5个数字的无序列表,范围是0-100,没有重复.我想检测3个数字是否连续且无间隔.
I'm working in Java. I have an unordered list of 5 numbers ranging from 0-100 with no repeats. I'd like to detect if 3 of the numbers are sequential with no gap.
示例:
[9,12,13,11,10] true
[17,1,2,3,5] true
[19,22,23,27,55] false
至于我尝试过的,什么都没有.如果我现在要编写它,我可能会采用最幼稚的方法来对数字进行排序,然后迭代检查序列是否存在.
As for what I've tried, nothing yet. If I were to write it now, I would probably go with the most naive approach of ordering the numbers, then iteratively checking if a sequence exists.
推荐答案
int sequenceMin(int[] set) {
int[] arr = Arrays.copy(set);
Arrays.sort(arr);
for (int i = 0; i < arr.length - 3 + 1; ++i) {
if (arr[i] == arr[i + 2] - 2) {
return arr[i];
}
}
return -1;
}
这将对数组进行排序,并使用上面的if语句查找所需的序列,并返回第一个值.
This sorts the array and looks for the desired sequence using the if-statement above, returning the first value.
不进行排序:
Without sorting:
(@ Pace提到了不进行排序的愿望.)在有限范围内可以使用有效的布尔数组" BitSet. nextSetBit
的迭代速度很快.
(@Pace mentioned the wish for non-sorting.) A limited range can use an efficient "boolean array", BitSet. The iteration with nextSetBit
is fast.
int[] arr = {9,12,13,11,10};
BitSet numbers = new BitSet(101);
for (int no : arr) {
numbers.set(no);
}
int sequenceCount = 0;
int last = -10;
for (int i = numbers.nextSetBit(0); i >= 0; i = numbers.nextSetBit(i+1)) {
if (sequenceCount == 0 || i - last > 1) {
sequenceCount = 1;
} else {
sequenceCount++;
if (sequenceCount >= 3) {
System.out.println("Sequence start: " + (last - 1));
break;
}
}
last = i;
}
System.out.println("Done");
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