#1480. Running Sum of 1d Array

Given an array nums.

We define a running sum of an array as runningSum[i] = sum(nums[0]…nums[i]).

Return the running sum of nums.

這是道簡單的題目 第一個想法是 foreach 就好 但要略過 index = 0 或是去判斷

判斷的所需秒數 12ms

` class Solution {

/**
 * @param Integer[] $nums
 * @return Integer[]
 */
function runningSum($nums) {
    $ans = [];
    foreach($nums as $index => $num){
        if($index === 0){
            $ans[$index] = $num;
        }else{
            $ans[$index] = $ans[$index - 1] + $num;
        }
    }
    return $ans;
}

} `

for迴圈略過為 4ms

` class Solution {

/**
 * @param Integer[] $nums
 * @return Integer[]
 */
function runningSum($nums) {
    $result[] = $nums[0];
    for ($i = 1; $i < count($nums); $i++) {
        $result[$i] = $nums[$i] + $result[$i - 1];
    }
    
    return $result;
}

} `