假设我有一个包含值[1,2,3,6,7]的数组。
如何检查数组以查看它是否包含3个连续数字。例如，上面的数组保持[1,2,3]，所以这将在我的函数中返回false。
var currentElement = null;
var counter = 0;
//check if the array contains 3 or more consecutive numbers:
for (var i = 0; i < bookedAppArray.length; i++) {
if ((bookedAppArray[i] != currentElement) && (bookedAppArray[i] === bookedAppArray[i - 1] + 1)) {
if (counter > 2) {
return true;
}
currentElement = bookedAppArray[i];
counter++;
} else {
counter = 1;
}
}
if(counter > 2){
return true;
} else{
return false;
}
答案 0 :(得分：1)
此解决方案
2，1 1
function consecutive(array) {
var i = 2, d;
while (i < array.length) {
d = array[i - 1] - array[i - 2];
if (Math.abs(d) === 1 && d === array[i] - array[i - 1]) {
return false;
}
i++;
}
return true;
}
document.write(consecutive([1]) + '<br>'); // true
document.write(consecutive([2, 4, 6]) + '<br>'); // true
document.write(consecutive([9, 8, 7]) + '<br>'); // false
document.write(consecutive([1, 2, 3, 6, 7]) + '<br>'); // false
document.write(consecutive([1, 2, 3, 4, 5]) + '<br>'); // false
答案 1 :(得分：0)
从逻辑上思考，这应该像迭代数组一样简单，只需检查当前数据之前的两个索引。
只需将1添加到上一个索引，将2添加到该索引之前的那个，它们应该都是相同的，就像这样
function hasThree(arr) {
var res = false;
arr.forEach(function(item, index) {
var l1 = arr[index - 1], // get previous
l2 = arr[index - 2]; // get the one before the previous
if ( l1 && l2 ) { // if two previous exist
// add 1, and then 2, and see if all are equal
if ( item === l1 + 1 && item === l2 + 2 ) res = true;
}
});
return res;
}
答案 2 :(得分：0)
有趣的问题。这是我的尝试。
#include <opencv2/core/core.hpp>
#include <opencv2/imgproc/imgproc.hpp>
#include <tiff.h>
#include <tiffio.h>
#include <stdint.h>
#include <string>
#include <vector>
#include <stdexcept>
#include <cstdio>
#include <cassert>
// just tiny exception generator
void except(bool condition, const std::string &message = "")
{
if (!condition)
throw std::runtime_error("Error: " + message);
}
bool write_fax(
const std::string &name, const cv::Mat &src, uint8_t threshold = 150)
{
cv::Mat image;
if (src.channels() == 3)
cv::cvtColor(src, image, CV_BGR2GRAY);
else if (src.channels() == 4)
cv::cvtColor(src, image, CV_BGRA2GRAY);
else
src.copyTo(image);
if (image.depth() != CV_8U)
{
cv::Mat tmp;
image.convertTo(tmp, CV_8U);
std::swap(image, tmp);
}
int width = image.cols;
int height = image.rows;
// do NOT put "wb" as the mode, because the b means "big endian" mode, not "binary" mode.
// http://www.remotesensing.org/libtiff/man/TIFFOpen.3tiff.html
TIFF* pTiffHandle = TIFFOpen(name.c_str(), "w");
if (!pTiffHandle)
{
printf("can't open TIFF descriptor\n");
return false;
}
try
{
except(TIFFSetField(pTiffHandle, TIFFTAG_IMAGEWIDTH, width), "width");
except(TIFFSetField(pTiffHandle, TIFFTAG_IMAGELENGTH, height), "length");
except(TIFFSetField(pTiffHandle, TIFFTAG_BITSPERSAMPLE, 1), "bits per sample");
except(TIFFSetField(pTiffHandle, TIFFTAG_SAMPLESPERPIXEL, 1), "samples per pixel");
except(TIFFSetField(pTiffHandle, TIFFTAG_ROWSPERSTRIP, 1), "rows per strip");
except(TIFFSetField(pTiffHandle, TIFFTAG_COMPRESSION, COMPRESSION_CCITTFAX4), "compression");
except(TIFFSetField(pTiffHandle, TIFFTAG_PHOTOMETRIC, PHOTOMETRIC_MINISWHITE), "photometric");
except(TIFFSetField(pTiffHandle, TIFFTAG_FILLORDER, FILLORDER_MSB2LSB), "photometric");
except(TIFFSetField(pTiffHandle, TIFFTAG_PLANARCONFIG, PLANARCONFIG_CONTIG), "planar config");
//except(TIFFSetField(pTiffHandle, TIFFTAG_PREDICTOR, predictor), "predictor");
//except(TIFFSetField(pTiffHandle, TIFFTAG_STRIPOFFSETS, strip_offsets), "strip offsets");
// not necessary
except(TIFFSetField(pTiffHandle, TIFFTAG_XRESOLUTION, 200.0), "res x");
except(TIFFSetField(pTiffHandle, TIFFTAG_YRESOLUTION, 200.0), "res y");
except(TIFFSetField(pTiffHandle, TIFFTAG_RESOLUTIONUNIT, RESUNIT_INCH), "res unit");
std::vector<uchar> _buffer(width / 8 + 8, 0);
uchar* buffer = &_buffer[0];
int bytes = int(width / 8.0 + 0.5);
for (int y = 0; y < height; ++y)
{
uint8_t *src_row = image.ptr(y);
for (int x = 0; x < width; ++x, ++src_row)
{
uint8_t eight_pixels = buffer[x / 8];
eight_pixels = eight_pixels << 1;
if (*src_row < threshold)
eight_pixels = eight_pixels | 1; //
buffer[x / 8] = eight_pixels;
}
// for the some reason writeEncodedStrip doesn't work
// except(TIFFWriteEncodedStrip(pTiffHandle, y, buffer, bytes) != -1, "write scanline");
except(TIFFWriteScanline(pTiffHandle, buffer, y, bytes) != -1, "write scanline");
}
}
catch (const std::runtime_error &e)
{
printf("TIFF writing: %s\n", e.what());
TIFFClose(pTiffHandle);
return false;
}
TIFFClose(pTiffHandle);
return true;
}
答案 3 :(得分：0)
这里是一个。只需确保对数组进行排序
function checkIfConsecutive(Arr) {
let isCnsc = true;
for (st in Arr ) {
if ( Arr[parseInt(st)+1]-Arr[parseInt(st)] > 1 && !isNaN(Arr[parseInt(st)+1]-Arr[parseInt(st)] )) {
isCnsc = false;
}
}
return isCnsc;
}
答案 4 :(得分：-1)
尝试使用Array.prototype.some()，Array.prototype.filter()
var arr1 = [1, 2, 3, 9, 8, 7];
var arr2 = [1, 2, "a", 3];
var check = function(a) {
// for each element in array `a`
return !a.some(function(item, index) {
// slice next three elements, including current element `item` from `a` array
var next = a.slice(index, 3);
console.log(next);
// if next three items in array `a` are type `Number`
// return `false`, else return `true`
return next.filter(Number).length === 3 ? true : false
})
};
// each item in `arr1` returns `false` , where item is followed
// by two numbers
// `arr2` returns `true` , where item is not followed by two numbers
console.log(check([1,2,3]), check(arr1), check(arr2)) // `false`, `false`, `true`
或者，使用for循环，Array.prototype.every()
var arr1 = [1, 2, 3, 9, 8, 7];
var arr2 = [1, 2, "a", 3];
var check = function(a) {
var res;
for (var i = 0; i < a.length; i++) {
// if `a[i]` is followed by two numbers, return `false`
if (a.slice(i, 3).every(function(n) {
return typeof n === "number"
})) {
res = false;
break;
}
// if `a[i]` is not followed by two numbers, return `true`
else {
res = true;
break;
}
}
return res
}
console.log(check([1,2,3]), check(arr1), check(arr2)) // `false`, `false` , `true`