关于快速选择功能

Question

首先，感谢您的帮助！我的问题是我试图找到最大的第k个测试用例，但有些测试用例失败了。如果参数是1, 1 ,1并且k=1，它将命中未定义的基本情况。我不知道为什么它会命中基本情况。再次，非常感谢你！如果你们需要更多信息，请告诉我！

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function kth_largest_in_an_array(numbers, k) {
    return quickSort3(numbers, k, 0, numbers.length-1);
}

const quickSort3 = (numbers, k, start, end) => {
    if (start >= end) {
        console.log('base case')
        return;
    }
    
    //remember random index
    console.log('before: ', numbers)
    let randomIdx =  Math.floor(start + Math.random() * (end - start));
    let partitionedIdx = partition(numbers, start, end, randomIdx);
    console.log('randomIdx: ',randomIdx)
    console.log('partitionedIdx: ',partitionedIdx)
    console.log('start: ', start, 'end: ', end)
    console.log(numbers.slice(start, end+1))
    console.log(numbers)
    console.log('number.length: ',numbers.length)
    //numbers.length = 5, 5-2 == 3
    // console.log('numbers.length - k: ', numbers.length - k)
    if (numbers.length - k === partitionedIdx) {
        console.log('imin: ', numbers[partitionedIdx])
        return numbers[partitionedIdx];
    }
    else if (partitionedIdx < numbers.length - k) {
        return quickSort3(numbers, k, partitionedIdx+1, end);
    }
    else {
        return quickSort3(numbers, k, start, partitionedIdx-1)
    }
}

const partition = (numbers, start, end, randomIdx) => {

    [numbers[start], numbers[randomIdx]] = [numbers[randomIdx], numbers[start]];
    let pivot = numbers[start];
    
    let i = start;
    for (let j = i+1; j<=end; j++) {
        if (pivot > numbers[j]) {
            i++
            [numbers[i], numbers[j]] = [numbers[j], numbers[i]];
        }
    }
    [numbers[start], numbers[i]] = [numbers[i], numbers[start]];
    
    return i;
}
console.log(kth_largest_in_an_array([1,1,1], 1))
// console.log(kth_largest_in_an_array([4, 1, 2, 2, 3, 4], 2))
// console.log(kth_largest_in_an_array([5, 1, 10, 3, 2], 2))

​

Answer 1

你在找这样的东西吗？

​

const t = (arr, kth) => {
  const a = arr.sort((a, b) => b - a);
  return a[kth];
}

console.log(t([1,1,1], 1))
console.log(t([4, 1, 2, 2, 3, 4], 2))
console.log(t([5, 1, 10, 3, 2], 2))

​