JS 递归相关问题 about frontend-interview HOT 5 OPEN

 function f() {    
       var arr = Array.from(arguments);
       let before = arr.slice(0, arr.length - 1);
       let n = arr[arr.length - 1];
       for (let i = 1; i < n; i++) {
            f(...before, i, n - i);
       }
       console.log(arr);
  }

from frontend-interview.

//减治法，O(n2)
var resolve = (function (){
    //闭包，缓存结果到table中
    var table = [
        [[]],
    ];
    /*初始表格，省去递归的边界条件检查;
    table[0]=[[]]
    table[1]=[[1]]
    */
    function fn(n){
        //查表，如果表中不存在
        if(!table[n]){
            var arr = [];
            /*
            f(1) = 0^f(1)=[[1]]
            f(2) = 1^f(1) U 0^f(2) = [[1,1],[2]]
            对于f(n),其结果为 [n]组合f(0) + [n-1]组合f(1) + [n-2]组合f(2) + ... + [1]组合f(n-1)
            */
            for(var i=n;i>=1;i--){
                fn(n-i).forEach(sub=>arr.push([i].concat(sub)))//将f(n-i)的每一项和i组合
            }
            table[n] = arr;//缓存结果
        }
        return table[n]
    }
    return fn;
})()

from frontend-interview.

@ucev 请问下可以写下思路吗？这算法写的太好了

from frontend-interview.

这是一个循环递归调用，首先
function f(n) {
if (n === 1) return [[1]];
var rs = [[n]];
for (var i = n - 1; i > 0; i--) {
rs = rs.concat(f(n - i).map(item => [i].concat(item)))
}
return rs
}
由于本题是输出结果，那就只有把需要输出的信息带过去
function f() {
var arr = Array.from(arguments);
console.log(arr)
var len = arr.length
var fRecursion = arr[0] //用于循环
var before = arr.slice(1, len)
for (var i = fRecursion - 1; i > 0; i--) {
f(i, ...before, fRecursion-i)
}
}

from frontend-interview.

function splitNum(num) {
let res =[]
help(0, [], num)
function help(cur, path, num) {
if (cur >= num) {
cur === num && res.push(path)
return
}
for (let i = 1; i <= num; i++) {
path.push(i)
cur += i
help(cur, path.slice(), num)
console.log(cur, path)
path.pop()
cur -=i
}
}
return res

}

from frontend-interview.