题目描述补充

给定一个整数数组nums和一个目标值target，请你在该数组中找出两个数，使得它们的和与目标值target的差的绝对值最小。返回这两个数的索引值（索引值从0开始），如果有多个答案存在，则返回任意一对即可。

示例

输入：

• nums = [2, 7, 11, 15]
• target = 9

输出：

• [0, 1] （因为2 + 7 = 9，是数组中两数之和最接近目标值9的组合，且索引为[0, 1]）

PHP 示例代码

function twoSumClosest($nums, $target) {
    $n = count($nums);
    $closestDiff = PHP_INT_MAX;
    $result = [-1, -1];

    for ($i = 0; $i < $n; $i++) {
        for ($j = $i + 1; $j < $n; $j++) {
            $diff = abs($nums[$i] + $nums[$j] - $target);
            if ($diff < $closestDiff) {
                $closestDiff = $diff;
                $result = [$i, $j];
            }
        }
    }

    return $result;
}

// 示例用法
$nums = [2, 7, 11, 15];
$target = 9;
$result = twoSumClosest($nums, $target);
print_r($result);

Python 示例代码

def twoSumClosest(nums, target):
    n = len(nums)
    closest_diff = float('inf')
    result = [-1, -1]

    for i in range(n):
        for j in range(i + 1, n):
            diff = abs(nums[i] + nums[j] - target)
            if diff < closest_diff:
                closest_diff = diff
                result = [i, j]

    return result

# 示例用法
nums = [2, 7, 11, 15]
target = 9
result = twoSumClosest(nums, target)
print(result)

JavaScript 示例代码

function twoSumClosest(nums, target) {
    let n = nums.length;
    let closestDiff = Infinity;
    let result = [-1, -1];

    for (let i = 0; i < n; i++) {
        for (let j = i + 1; j < n; j++) {
            let diff = Math.abs(nums[i] + nums[j] - target);
            if (diff < closestDiff) {
                closestDiff = diff;
                result = [i, j];
            }
        }
    }

    return result;
}

// 示例用法
const nums = [2, 7, 11, 15];
const target = 9;
const result = twoSumClosest(nums, target);
console.log(result);

码小课提示

以上代码示例通过双重循环遍历数组，计算每两个数的和与目标值的差的绝对值，并更新最小差值及其对应的索引。虽然这种方法在理解上较为直观，但时间复杂度较高（O(n^2)），对于大数据集可能不够高效。码小课网站中分享了更多关于算法优化的内容，如哈希表等数据结构的应用，可以显著提高解决此类问题的效率。