问题描述
我正在解决 javascript 中的一个问题,我应该编写一个函数,该函数接受一个整数数组和一个“偶数”或“奇数”字符串。该函数将计算连续出现 4 个偶数或 4 个奇数的次数。
例如:
quadruples([3,2,4,8,5],'even') // 1
quadruples([2,6,10,'even') // 2
quadruples([2,'odd') // 0
到目前为止,这是我所在的位置:
我想我需要运行一个 for 循环,然后使用一个 % 运算符,但我被困在从这里开始的地方。
感谢任何帮助!
解决方法
为此您需要使用局部和全局变量进行动态编程: [2,4,6,8,10,5]
- 2 - 偶数,count 为 1,totalCount 为 0
- 4 - 偶数,count 为 2,totalCount 为 0
- 6 - 偶数,count 为 3,totalCount 为 0
- 8 - 偶数,count 为 4,totalCount 为 0
- 10 - 偶数,count 为 5,totalCount 为 0
- 5 - 奇数,计数为 5,将 totalCount 增加 5 - 4 + 1 = 2,将计数重置为 0
const quadruples = (givenArray,evenOrOdd) => {
// never hardcode `magic numbers`,create constants for them
const sequenceLength = 4
// based on evenOrOdd calculating what the division by 2
// will be if it is even,then 0,if it is odd,then 1
const rest = evenOrOdd === 'even' ? 0 : 1
// this will hold the total count of quadruples
let totalCount = 0
// this is the local count of contiguous elements
let count = 0
// looping over the array
for (let i = 0; i <= givenArray.length; i += 1) {
const el = givenArray[i]
// if the element is not what we want
if (i === givenArray.length || el % 2 !== rest) {
// if the count is 4 or more,we add to totalCount the count
// minus 4 and plus 1,meaning that if we have 4,it's 1 quadruple,// if it is 5,then it's 2 quadruples,etc.
// Otherwise (count is less than 4) we add 0 (nothing)
totalCount += count >= sequenceLength ? count - sequenceLength + 1 : 0
// resetting the count to zero as we encountered the opposite
// of what we are looking for (even/odd)
count = 0
// if the element is what we need (even or odd)
} else {
// increasing the count of how many we've seen by far
count += 1
}
}
// returning totalCount of quadruples
return totalCount
}
console.log(quadruples([1,3,5,7,9,11],'odd')) // 3
console.log(quadruples([3,2,5],'even')) // 1
console.log(quadruples([2,'even')) // 2
console.log(quadruples([2,'odd')) // 0
我写了这个递归。
console.log(quadruples([3,'even')); // 1
console.log(quadruples([2,'even')); // 2
console.log(quadruples([2,'odd')); // 0
console.log(quadruples([5,4],'even')); // 4
function quadruples(givenArray,evenOrOdd) {
const maxSequence = 4;
let result = 0;
if (givenArray.length < maxSequence)
return 0;
for (let i = 0; i < maxSequence; i++) {
if (givenArray[i] % 2 != (evenOrOdd == "even" ? 0 : 1)) {
givenArray = givenArray.slice(i + 1);
return (result += quadruples(givenArray,evenOrOdd));
}
}
result++;
givenArray = givenArray.slice(1);
return (result += quadruples(givenArray,evenOrOdd));
}