【Leetcode】python – [494] Target Sum 個人解法筆記

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題目出處

494. Target Sum

難度

medium

個人範例程式碼

class Solution:
    def findTargetSumWays(self, nums: List[int], target: int) -> int:

        memo = {}
        return self.dfs(nums, 0, target, memo)

    def dfs(self, nums, start_idx, target, memo):
        # end of recursion
        if start_idx >= len(nums):
            if target == 0:
                return 1
            else:
                return 0

        # memoization
        if (start_idx, target) in memo:
            return memo[(start_idx, target)]

        # define and split
        ans = self.dfs(nums, start_idx+1, target-nums[start_idx], memo) + self.dfs(nums, start_idx+1, target+nums[start_idx], memo)

        # memoization
        memo[(start_idx, target)] = ans

        return ans

最近在練習程式碼本身就可以自解釋的 Coding style,可以嘗試直接閱讀程式碼理解

算法說明

仔細看發現是很單純的「2 指數」的題目,
每一次我們都只要多延伸出「+」或「-」的新情況,

memoization

純 DFS 會 TLE,需要搭配 DP 的記憶化搜索,
我們紀錄的是,(目前 target 的值,目前的 idx),
當進行一樣的 target 與同 idx 搜尋時,我們可以直接查表。

input handling

在 dfs 內處理

Boundary conditions

在 dfs 內控制範圍,我們在「搜尋 start_idx >= nums 總長」的情況下停止遞迴。

Reference

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