问题
I am trying to get the range of numbers using recursion. Can someone explain to me why it isn't working?
function range(x,y){
var results = [];
if(x === y){
return results;
}
return results.push(range(x + 1,y));
}
range(1,5);
回答1:
Try this:
function rangeOfNumbers(startNum, endNum) {
if (startNum - endNum === 0) {
return [startNum];
} else {
const numbers = rangeOfNumbers(startNum + 1, endNum);
numbers.unshift(startNum);
return numbers;
}
};
回答2:
The beauty of recursion is that you don't need local variables (var results
). You just pass state as arguments to each recursive iteration:
const concat = (xs, y) => xs.concat(y);
const range = (x, y) => {
const rec = (x, y, acc) => x < y ? rec(x + 1, y, concat(acc, x)) : acc;
return rec(x, y, []);
}
ES5 version in case you aren't familiar with the arrow syntax:
function concat(xs, y) {
return xs.concat(y);
}
function range(x, y) {
function rec(x, y, acc) {
return x < y ? rec(x + 1, y, concat(acc, x)) : acc;
}
return rec(x, y, []);
}
That isn't the most elegant solution though!
With recursion we can simply build up the stack with each recursive call. Each stack frame contains a computed partial result. Then we just need to unwind the stack and attach each partial result to an array:
const range = (x, y) => x < y ? [x].concat(range(x + 1, y)) : [];
Or more functional:
const concat = (xs, y) => xs.concat(y);
const range = (x, y) => x < y ? concat([x], range(x + 1, y)) : [];
Note that concat([x], range(x + 1, y))
is the recursive case and []
the base case.
回答3:
Results will be always empty since you actually don't put anything in it.
What would work is this
function range(x,y){
var results = [];
if(x === y){
return results;
}
results.push(x);
return results.concat(range(x + 1,y));
}
range(1,5);
回答4:
Solution: Solved this recursion problem, which is taking 2 numbers as input and returning back the array which contains range of the numbers inclusive of the startNumber and EndNumber
Assumption-> end_num is always greater than start_num
function rangeOfNumbers(start_num, end_num) {
if(start_num!==end_num){
let numbers = rangeOfNumbers(start_num+1,end_num);
numbers.unshift(start_num);
return numbers;
}
else
return [start_num];
};
来源:https://stackoverflow.com/questions/38175342/getting-range-numbers-using-recursion-in-javascript