算法题：两数之和

1. Two Sum

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.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

一、两遍循环，暴力破解

代码如下

function twoSum($nums, $target) {
        for($i=0;$i<count($nums); $i++){
            for($j=$i+1; $j<count($nums); $j++){
                $sum = $nums[$i]+$nums[$j];
                if($target == $sum){
                    return array($i,$j);
                }
            }
        }
    }

时间复杂度 O(n^2 )

提交，结果执行时间1968 ms。。。。。

可以说龟速了

二、两遍hash

代码如下

function twoSum2(array $nums , $target)
{
    $res = [];
    $nums_match = [];
    foreach ($nums as $nums_k => $nums_v){
        if(!isset($nums_match[$target-$nums_v])){
            $nums_match[$target-$nums_v] = $nums_k;
        }
    }
    
    foreach ($nums as $nums_k => $nums_v){
        if (isset($nums_match[$nums_v]) && $nums_match[$nums_v] != $nums_k) {
            $res[] = $nums_k;
            $res[] = $nums_match[$nums_v];
            return $res;
        }
    }
}

时间复杂度O(n)

执行时间24 ms ，提升很大

三、一遍hash

这是在两边hash的基础上进行的优化

代码如下

function twoSum($nums, $target) {
        $nums_match = [];
        foreach ($nums as $nums_k => $nums_v){
            if((isset($nums_match[$target-$nums_v]))){
                return array($nums_match[$target-$nums_v],$nums_k);
            }
            $nums_match[$nums_v] = $nums_k;
        }

    }

时间复杂度O(n)

执行时间16 ms