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第 8 期(2019-05-15):求斐波那契数前N项 #10

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@wingmeng

来源:面试题
难度:★★

封装一个函数,接收1个整数参数 n,返回 斐波那契数列 前 n 项

/**
 * @param {number} n - 整数
 */
function fibonacci(n) {
  // 你的代码
}

参考答案:

function fibonacci(n) {
  var a = 0;
  var b = 1;
  var c = a + b;
  var arr = [c];

  for (var i = arr.length; i < n; i++) {
    arr.push(c);
    a = b;
    b = c;
    c = a + b;
  }

  return arr;
}

// 再来个使用递归+ES6实现的(原创)
function fibonacci(n) {
  let fn = i => i < 3 ? 1 : fn(i - 1) + fn(i - 2);
  return [...Array(n).keys()].map(it => fn(it + 1));
}

本期优秀回答者: @AMY-Y

Activity

  1. liwenkang commented on May 15, 2019

    @liwenkang
    function getFibonacciArray(n) {
        let result = [];
        let map = new Map();
        map.set(1, 1);
        map.set(2, 1);
    
        function fibonacci(n) {
            if (map.get(n)) {
                return map.get(n);
            } else {
                let result = fibonacci(n - 2) + fibonacci(n - 1);
                map.set(n, result);
                return result;
            }
        }
    
        fibonacci(n);
        return map.values();
    }
  2. AMY-Y commented on May 15, 2019

    @AMY-Y
    function feibonacci(n){
                if(Math.round(n)===n){
                    var arr=[];
                    var sum1=1;
                    var sum2=1;
                    if(n==1){
                        arr.push(sum1);
                    }else{
                        arr.push(sum1);
                        arr.push(sum2);
                        for(var i=2;i<n;i++){
                            var temp;
                            temp=sum1+sum2;
                            sum1=sum2;
                            sum2=temp;
                            arr.push(temp);
                        }
                    }                
                    return arr;
                }
            }
            feibonacci(8);
  3. ahao430 commented on May 16, 2019

    @ahao430
    function fibonacci(n) {
      if (n < 0) return []
    
      let res = []
      for (let i = 0; i < n; i++) {
        let num
        if (i === 0) num = 0
        else if (i === 1) num = 1
        else num = res[i - 1] + res[i - 2]
        res.push(num)
      }
      return res
    }
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