Viiisit [LeetCode] - 1480. Running Sum of 1d Array

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Problem - 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.

翻譯蒟蒻

計算 nums 陣列中每個元素從第一個元素開始到當前元素的和，並將這些和作為新的陣列返回。

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• Example 1:

  Input: nums = [1,2,3,4]
  Output: [1,3,6,10]
  Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].

• Example 2:

  Input: nums = [1,1,1,1,1]
  Output: [1,2,3,4,5]
  Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1].

• Example 3:

  Input: nums = [3,1,2,10,1]
  Output: [3,4,6,16,17]

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Solution - JavaScript

Solution 1:
藉由建立一個新的陣列來儲存加總後的結果。

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var runningSum = function(nums) {
  let arr = [];
  for (let i = 0; i < nums.length; i++) {
    let sum = 0;
    for (let j = 0; j <= i; j++) {
      sum += nums[j];
    }
    arr.push(sum);
  }
  return arr;
};

Solution 2:
加總之後直接改變陣列的值，作為新的陣列輸出。

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var runningSum = function(nums) {
  for (let i = 1; i < nums.length; i++) {
    nums[i] += nums[i - 1];
  }
  return nums;
};

Solution - Ruby

Solution 1:
先設定 sum 初始值為 0，透過 map 方法累加成為新的陣列。

def running_sum(nums)
  sum = 0
  nums.map { |num| sum += num }
end

Solution 2:
使用 map 與 with_index，計算索引 0 到當前索引 index 的所有數字的總和。

def running_sum(nums)
  nums.map.with_index do |num, index|
    nums[0..index].sum
  end
end

Solution - PHP

Solution 1:

function runningSum($nums) {
  for ($i = 1; $i < count($nums); $i++) {
      $nums[$i] += $nums[$i - 1];
  }
  return $nums;
}

Solution 2:

function runningSum($nums) {
  $sum = 0;
  $result = [];
  for ($i = 0; $i < count($nums); $i++) {
      $sum += $nums[$i];
      array_push($result, $sum);
  }
  return $result;
}

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