Question

我只能求和10。不能正确。我得到的是10位数字的总和（仅过滤7位奇数后）。我应该在哪里<= num？

function sumFibs(num) {
  var fib = [1, 1];
  for (var i = 2; i < num; i++) {
    var next = fib[i - 1] + fib[i - 2];
    var fibi = fib.push(next);
  }

  return fib.filter(function(a) {
      return (a % 2 != 0);
    })
    .reduce(function(a, z) {
      return a + z;
    })
}
console.log(sumFibs(10));

预期输出10，但得到99

Answer 1

将a < num添加到过滤器回调测试中，以便获得a % 2 && a < num

function sumFibs(num) {
	var fib = [0, 1];
	for (var i = 2; i < num; i++) {
		var next = fib[i - 1] + fib[i - 2];
		var fibi = fib.push(next);
	}
	return fib.filter(function (a) {
		return a % 2 && a < num;
	}).reduce(function (a, z) {
		return a + z;
	}, 0);
}

console.log(sumFibs(0))
console.log(sumFibs(1))
console.log(sumFibs(10))
console.log(sumFibs(9000))

如果只需要这些数字的总和，就不需要使用数组

function sumFibs(num) {
  if(num <= 1) return 0;
  var a = 0, b = 1, sum = a + b;
  while(true) {
    var next = a + b;
    if(next >= num) {
      break;
    }
    if(next % 2) {
      sum += next;
    }
    a = b;
    b = next;
  }
  return sum
}
console.log(sumFibs(0))
console.log(sumFibs(1))
console.log(sumFibs(10))
console.log(sumFibs(9000))

Answer 2

您需要更改循环条件。循环直到fib的最后一个值小于num

function sumFibs(num) {
  var fib = [1, 1];
  for (var i = 2; fib[fib.length - 1] < num; i++) {
    var next = fib[i - 1] + fib[i - 2];
    fib.push(next);
  }
   return fib
    .filter(x => !(x % 2))
    .reduce((ac,a) => ac + a,0)
}
console.log(sumFibs(10));

Answer 3

我想这就是你想要的：

function sumFibs(num) {
  var fib = [1, 1];
  var sum = 2;
  for (var i = 2; i < num; i++) {
    var next = fib[i - 1] + fib[i - 2];
    var fibi = fib.push(next);
    if (next<=num && next % 2 != 0)
      sum += next;
  }

  console.log("fib: " + fib);
  console.log("sum: " + sum);
  return sum;
}

console.log(sumFibs(10));

Answer 4

如果您想保留大部分代码，则可以添加if语句，一旦您的next数超出num，就可以中断循环：

function sumFibs(num) {
  var fib = [1, 1];
  for (var i = 2; i < num; i++) {
    var next = fib[i - 1] + fib[i - 2];

    if ( next > num ) { // not >= assuming you want to include your num
      break;
    }

    fib.push(next);
  }
  console.log({fib});
  
  return fib.filter(function(a) {
      return (a % 2 != 0);
    })
    .reduce(function(a, z) {
      return a + z;
    })
}
console.log(sumFibs(10));

Answer 5

您的代码正在将前N个斐波那契数相加。您似乎正在寻找总和为N的第一个奇数斐波那契数：

function oddFibsThatAddTo(target)
{
  let currentFib = 1;
  let lastFib = 1;
  let sum = 2;
  const outs = [1,1];
  while(sum < target)
  {
    let nextFib = currentFib + lastFib;
    if(nextFib % 2 == 1)
    {
      sum += nextFib;
      outs.push(nextFib);
    }
    lastFib = currentFib;
    currentFib = nextFib;
  }
  if(sum > target)
  {
    throw 'can\'t find perfect sequence';
  }
  return outs;
}

console.log(oddFibsThatAddTo(10))

Answer 6

function sumFibs(num) {
  var fib=[1,1];
  for(var i=2; i<num; i++){
  var next=fib[i-1]+fib[i-2];
  var fibi=fib.push(next);
}
return fib.filter(function(a){
  return(a%2!=0 && a<=num);
})
.reduce(function(a,z){
 return a+z;
})
}
console.log(sumFibs(10));

所有奇数斐波纳契数之和，最高为n

6 个答案: