PHP实现数组中两个数的和等于给定的目标值

算法：
1、以数组中的值为索引创建新的数组$tmp
2、求出目标值减去数组值得差值
3、判断该差值是否在\$tmp中。
php实现代码如下

/** * Given an array of integers, return indices of the two numbers such that they add up to a specific target. * You may assume that each input would have exactly one solution, and you may not use the same  * element twice. * Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. * @param  [type] $arr [description] * @param  [type] $sum [description] * @return [type]      [description] */function twoSum($arr, $sum) {    $len = count($arr);    if ($len < 2) return false;    $tmp = [];    $flag = [];    // 以数组的值为索引    for ($i = 0; $i < $len; $i++) {        $tmp[$arr[$i]] = $i;    }    // 判断差值是否在上述索引数组中    for ($j = 0; $j < $len; $j++) {        $minus = $sum - $arr[$j];        if ($minus < 0 || $sum <= $arr[$j]) continue;        if (isset($tmp[$minus]) && !isset($flag[$arr[$j]]) && !isset($flag[$minus])) {            echo '数组的索引值为[' . $tmp[$arr[$j]] . ',' . $tmp[$minus] . ']<br>';            // 如果有则将值置为1            $flag[$arr[$j]] = 1;        }    }}$arr = [2, 7, 11, 15];//$arr = [1,2,7,9,8,3,6,5,4,10];twoSum($arr, 9);

时间复杂度为O(n)。