需要一个从给定数组返回N个元素的函数,给定偏移量,但当偏移量大于数组长度时,它必须返回数组开头的元素。
接口
slice2(array, chunk, offset);
示例:
var array = [1,2,3,4,5];
slice2(array,2,2)
输出:[3,4]
slice2(array,2,4)
输出:[5,1]
slice2(array,3,4)
输出:[5,1,2]
答案 0 :(得分:7)
function slice2(array, chunk, offset) {
var subarray = [];
for (var i = 0; i<chunk; i++) {
var ind = (offset + i) % array.length;
subarray.push(array[ind]);
}
return subarray;
}
答案 1 :(得分:0)
内置的.slice()方法可能足以解决您的问题:
var slice = array.slice(1,3);
我设置了一个快速的jsfiddle,界面需要改变,因为它只允许出现
答案 2 :(得分:0)
对不起自己也不能帮忙:)。
以下函数可以处理更多场景,并且仅使用2个切片执行此操作。
// standard
slice2( [1,2,3,4,5], 3, 3 ); // [1, 4, 5]
// specify chunk as a negative number
slice2( [1,2,3,4,5], -3, 3 ); // [1, 2, 3]
// what if chunk is larger than array
slice2( [1,2,3,4,5], -7, 3 ); // [1, 2, 3, 4, 5]
// what if chunk is 0
slice2( [1,2,3,4,5], 0, 3 ); // []
function slice2( array, chunk, offset ) {
var start1, len = array.length, end1, start2, end2, result;
if( !chunk || !array ) return []; // if chunk or array resolve to falsy value.
if( Math.abs(chunk) >= len ) return array;
if( chunk < 0 ) {
end1 = offset;
leftover = offset + chunk;
start1 = leftover < 0 ? 0 : leftover;
start2 = leftover;
end2 = len;
} else {
start1 = offset;
leftover = ( offset + chunk ) - len;
end1 = leftover > 0 ? ( offset + chunk ) : len;
start2 = 0;
end2 = leftover;
}
result = array.slice( start1, end1 );
if( leftover ) result = array.slice( start2, end2 ).concat( result );
return result;
}
小提琴here
答案 3 :(得分:0)
试试这个:
function slice2(array, chunk, offset){
var end = offset + chunk,
out = array.slice(offset, end); // Get the chunk
if(array.length < end){ // If the chunk should wrap
out = out.concat(array.slice(0, end - array.length)); // Concatenate a the rest of the chunk, from the start of the array, to the output.
}
return out;
}
它不会手动循环遍历数组,只使用极少的计算。
答案 4 :(得分:0)
private static List sliceArray(List mainarray,int chunk,int offset){
int size = mainarray.size();
List<Integer> resultArray = new ArrayList<Integer>();
if(!mainarray.contains(chunk)){
return null;
}
int index = mainarray.indexOf(chunk);
int doOffset = size - index;
if(doOffset > offset){
for(int element = 1 ; element <= offset; element++ ){
resultArray.add(mainarray.get(index+1));
index++;
}
}else if(doOffset == 0){
for(int element = 0 ; element <= offset; element++ ){
element = mainarray.get(element);
resultArray.add(element);
}
}else{
int position =0;
for(int element = index ; element <= offset; element++ ){
int value = 0;
if(element < size-1){
value = mainarray.get(element+1);
}
else
{
value = mainarray.get(position);
position ++;
}
resultArray.add(value);
}
}
return resultArray;
// TODO Auto-generated method stub
}
答案 5 :(得分:0)
这是我所做的摘要。感谢@Andrei我刚刚修改了他的代码。因此,我想要的是一个函数,用于获取数组中的选定数字以及该数字之前和之后的x个数字。例如,[1,2,3,4,5,6,7,8]如果我选择2作为偏移量,将块选择为3,则输出将是:[8,1,2,2,3,4,5,6] 如果offset = 3,chunk = 3,op:[1,2,3,4,5,6,7]
function cyclicSlice(array, chunk, offset) {
var subarray = [];
for (var i = array.length; i>=(array.length-chunk); i--) {
var ind = (offset + i) % array.length;
subarray.push(array[ind]);
}
subarray = subarray.reverse();
const newOffset = offset + 1; // To avoid the repititon of the selected number
for (var i = 0; i<chunk; i++) {
var ind = ( newOffset + i) % array.length;
subarray.push(array[ind]);
}
return subarray;
}