generate the corresponding loop based on the given number. If 2 is given, a double cycle is generated, and 3 is a triple cycle
if 3 is given, then
for (let i = 0; i < arr.length; iPP) {
for (let j = 0; j < arr.length; jPP) {
for (let k = 0; k < arr.length; kPP) {
}
}
}
}Recursive
f(arr, 3, [])
function f(arr,k,temp){
if(k<=0){
let sum = 0
let r =[]
for(let i = 0; i < temp.length; iPP){
sum += temp[i]
r.push(temp[i])
}
if (sum === hz) {
result.push(r)
}
return
}
for (let i = 0; i < arr.length; iPP) {
temp.push(arr[i])
f(arr, k-1, temp)
temp.splice(temp.length-1, 1)
}
}
this kind of demand can usually be solved through negotiations, so let n count and then n = 0 jump out of the box. It is estimated that it will take some tricks and tricks to force loop to solve the problem.
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there is a set of numbers, for example, 3, requires that the sum equal to 3 is selected from this group of numbers (3 added numbers). For example, the three added numbers cannot be the same (for example, 1: 1, 1: 1), and there is no order requirement fo...