<ruby id="bdb3f"></ruby>

    <p id="bdb3f"><cite id="bdb3f"></cite></p>

      <p id="bdb3f"><cite id="bdb3f"><th id="bdb3f"></th></cite></p><p id="bdb3f"></p>
        <p id="bdb3f"><cite id="bdb3f"></cite></p>

          <pre id="bdb3f"></pre>
          <pre id="bdb3f"><del id="bdb3f"><thead id="bdb3f"></thead></del></pre>

          <ruby id="bdb3f"><mark id="bdb3f"></mark></ruby><ruby id="bdb3f"></ruby>
          <pre id="bdb3f"><pre id="bdb3f"><mark id="bdb3f"></mark></pre></pre><output id="bdb3f"></output><p id="bdb3f"></p><p id="bdb3f"></p>

          <pre id="bdb3f"><del id="bdb3f"><progress id="bdb3f"></progress></del></pre>

                <ruby id="bdb3f"></ruby>

                ??一站式輕松地調用各大LLM模型接口,支持GPT4、智譜、豆包、星火、月之暗面及文生圖、文生視頻 廣告
                # 0322. 零錢兌換 ## 題目地址(322. 零錢兌換) <https://leetcode-cn.com/problems/coin-change/> ## 題目描述 ``` <pre class="calibre18">``` 給定不同面額的硬幣 coins 和一個總金額 amount。編寫一個函數來計算可以湊成總金額所需的最少的硬幣個數。如果沒有任何一種硬幣組合能組成總金額,返回 -1。 你可以認為每種硬幣的數量是無限的。 示例 1: 輸入:coins = [1, 2, 5], amount = 11 輸出:3 解釋:11 = 5 + 5 + 1 示例 2: 輸入:coins = [2], amount = 3 輸出:-1 示例 3: 輸入:coins = [1], amount = 0 輸出:0 示例 4: 輸入:coins = [1], amount = 1 輸出:1 示例 5: 輸入:coins = [1], amount = 2 輸出:2 提示: 1 <= coins.length <= 12 1 <= coins[i] <= 231 - 1 0 <= amount <= 104 ``` ``` ## 前置知識 - 貪心算法 - [動態規劃](https://github.com/azl397985856/leetcode/blob/master/thinkings/dynamic-programming.md) ## 公司 - 騰訊 - 百度 - 字節 - 阿里巴巴(盒馬生鮮) ## 崗位信息 - 阿里巴巴(盒馬生鮮):前端技術二面 ## 思路 ## 思路 假如我們把 coin 逆序排列,然后逐個取,取到剛好不大于 amout,依次類推。 ``` <pre class="calibre18">``` eg: 對于 [1,2,5] 組成 11 塊 - 排序[5,2,1] - 取第一個5, 更新amout 為 11 - 5 = 6 (1??) 6 > 5 繼續更新 為 6 - 5 = 1 (2??) 1 < 5 退出 - 取第二個2 1 < 2 退出 - 取最后一個元素,也就是1 1 === 1 更新為 1 - 1 = 0 (3??) - amout 為 0 退出 因此結果是 3 ``` ``` 熟悉貪心算法的同學應該已經注意到了,這就是貪心算法,貪心算法更 amount 盡快地變得更小。 `經驗表明,貪心策略是正確的`。 注意,我說的是經驗表明, 貪心算法也有可能出錯。 就拿這道題目來說, 他也是不正確的! 比如 `coins = [1, 5, 11] amout = 15`, 因此這種做法有時候不靠譜,我們還是采用靠譜的做法. 如果我們暴力求解,對于所有的組合都計算一遍,然后比較, 那么這樣的復雜度是 2 的 n 次方(這個可以通過數學公式證明,這里不想啰嗦了), 這個是不可以接受的。那么我們是否可以動態規劃解決呢?答案是可以,原因就是可以劃分為子問題,子問題可以推導出原問題 對于動態規劃我們可以先畫一個二維表,然后觀察,其是否可以用一維表代替。 關于動態規劃為什么要畫表,我已經在[這篇文章](dynamic-programming.html)解釋了 比較容易想到的是二維數組: ``` <pre class="calibre18">``` <span class="hljs-class"><span class="hljs-keyword">class</span> <span class="hljs-title">Solution</span>:</span> <span class="hljs-function"><span class="hljs-keyword">def</span> <span class="hljs-title">coinChange</span><span class="hljs-params">(self, coins: List[int], amount: int)</span> -> int:</span> <span class="hljs-keyword">if</span> amount < <span class="hljs-params">0</span>: <span class="hljs-keyword">return</span> - <span class="hljs-params">1</span> dp = [[amount + <span class="hljs-params">1</span> <span class="hljs-keyword">for</span> _ <span class="hljs-keyword">in</span> range(len(coins) + <span class="hljs-params">1</span>)] <span class="hljs-keyword">for</span> _ <span class="hljs-keyword">in</span> range(amount + <span class="hljs-params">1</span>)] <span class="hljs-title"># 初始化第一行為0,其他為最大值(也就是amount + 1)</span> <span class="hljs-keyword">for</span> j <span class="hljs-keyword">in</span> range(len(coins) + <span class="hljs-params">1</span>): dp[<span class="hljs-params">0</span>][j] = <span class="hljs-params">0</span> <span class="hljs-keyword">for</span> i <span class="hljs-keyword">in</span> range(<span class="hljs-params">1</span>, amount + <span class="hljs-params">1</span>): <span class="hljs-keyword">for</span> j <span class="hljs-keyword">in</span> range(<span class="hljs-params">1</span>, len(coins) + <span class="hljs-params">1</span>): <span class="hljs-keyword">if</span> i - coins[j - <span class="hljs-params">1</span>] >= <span class="hljs-params">0</span>: dp[i][j] = min( dp[i][j - <span class="hljs-params">1</span>], dp[i - coins[j - <span class="hljs-params">1</span>]][j] + <span class="hljs-params">1</span>) <span class="hljs-keyword">else</span>: dp[i][j] = dp[i][j - <span class="hljs-params">1</span>] <span class="hljs-keyword">return</span> <span class="hljs-params">-1</span> <span class="hljs-keyword">if</span> dp[<span class="hljs-params">-1</span>][<span class="hljs-params">-1</span>] == amount + <span class="hljs-params">1</span> <span class="hljs-keyword">else</span> dp[<span class="hljs-params">-1</span>][<span class="hljs-params">-1</span>] ``` ``` **復雜度分析** - 時間復雜度:O(amonut?len(coins))O(amonut \* len(coins))O(amonut?len(coins)) - 空間復雜度:O(amount?len(coins))O(amount \* len(coins))O(amount?len(coins)) dp\[i\]\[j\] 依賴于`dp[i][j - 1]`和 `dp[i - coins[j - 1]][j] + 1)` 這是一個優化的信號,我們可以將其優化到一維,具體見下方。 ## 關鍵點解析 - 動態規劃 - 子問題 用 dp\[i\] 來表示組成 i 塊錢,需要最少的硬幣數,那么 1. 第 j 個硬幣我可以選擇不拿 這個時候, 硬幣數 = dp\[i\] 2. 第 j 個硬幣我可以選擇拿 這個時候, 硬幣數 = dp\[i - coins\[j\]\] + 1 3. 和背包問題不同, 硬幣是可以拿任意個 4. 對于每一個 dp\[i\] 我們都選擇遍歷一遍 coin, 不斷更新 dp\[i\] ## 代碼 - 語言支持:JS,C++,Python3 JavaScript Code: ``` <pre class="calibre18">``` <span class="hljs-keyword">var</span> coinChange = <span class="hljs-function"><span class="hljs-keyword">function</span> (<span class="hljs-params">coins, amount</span>) </span>{ <span class="hljs-keyword">if</span> (amount === <span class="hljs-params">0</span>) { <span class="hljs-keyword">return</span> <span class="hljs-params">0</span>; } <span class="hljs-keyword">const</span> dp = <span class="hljs-params">Array</span>(amount + <span class="hljs-params">1</span>).fill(<span class="hljs-params">Number</span>.MAX_VALUE); dp[<span class="hljs-params">0</span>] = <span class="hljs-params">0</span>; <span class="hljs-keyword">for</span> (<span class="hljs-keyword">let</span> i = <span class="hljs-params">1</span>; i < dp.length; i++) { <span class="hljs-keyword">for</span> (<span class="hljs-keyword">let</span> j = <span class="hljs-params">0</span>; j < coins.length; j++) { <span class="hljs-keyword">if</span> (i - coins[j] >= <span class="hljs-params">0</span>) { dp[i] = <span class="hljs-params">Math</span>.min(dp[i], dp[i - coins[j]] + <span class="hljs-params">1</span>); } } } <span class="hljs-keyword">return</span> dp[dp.length - <span class="hljs-params">1</span>] === <span class="hljs-params">Number</span>.MAX_VALUE ? <span class="hljs-params">-1</span> : dp[dp.length - <span class="hljs-params">1</span>]; }; ``` ``` C++ Code: > C++中采用 INT\_MAX,因此判斷時需要加上`dp[a - coin] < INT_MAX`以防止溢出 ``` <pre class="calibre18">``` <span class="hljs-keyword">class</span> Solution { <span class="hljs-keyword">public</span>: <span class="hljs-function"><span class="hljs-keyword">int</span> <span class="hljs-title">coinChange</span><span class="hljs-params">(<span class="hljs-params">vector</span><<span class="hljs-keyword">int</span>>& coins, <span class="hljs-keyword">int</span> amount)</span> </span>{ <span class="hljs-keyword">auto</span> dp = <span class="hljs-params">vector</span><<span class="hljs-keyword">int</span>>(amount + <span class="hljs-params">1</span>, INT_MAX); dp[<span class="hljs-params">0</span>] = <span class="hljs-params">0</span>; <span class="hljs-keyword">for</span> (<span class="hljs-keyword">auto</span> a = <span class="hljs-params">1</span>; a <= amount; ++a) { <span class="hljs-keyword">for</span> (<span class="hljs-keyword">const</span> <span class="hljs-keyword">auto</span> & coin : coins) { <span class="hljs-keyword">if</span> (a >= coin && dp[a - coin] < INT_MAX) { dp[a] = min(dp[a], dp[a-coin] + <span class="hljs-params">1</span>); } } } <span class="hljs-keyword">return</span> dp[amount] == INT_MAX ? <span class="hljs-params">-1</span> : dp[amount]; } }; ``` ``` Python3 Code: ``` <pre class="calibre18">``` <span class="hljs-class"><span class="hljs-keyword">class</span> <span class="hljs-title">Solution</span>:</span> <span class="hljs-function"><span class="hljs-keyword">def</span> <span class="hljs-title">coinChange</span><span class="hljs-params">(self, coins: List[int], amount: int)</span> -> int:</span> dp = [amount + <span class="hljs-params">1</span>] * (amount + <span class="hljs-params">1</span>) dp[<span class="hljs-params">0</span>] = <span class="hljs-params">0</span> <span class="hljs-keyword">for</span> i <span class="hljs-keyword">in</span> range(<span class="hljs-params">1</span>, amount + <span class="hljs-params">1</span>): <span class="hljs-keyword">for</span> j <span class="hljs-keyword">in</span> range(len(coins)): <span class="hljs-keyword">if</span> i >= coins[j]: dp[i] = min(dp[i], dp[i - coins[j]] + <span class="hljs-params">1</span>) <span class="hljs-keyword">return</span> <span class="hljs-params">-1</span> <span class="hljs-keyword">if</span> dp[<span class="hljs-params">-1</span>] == amount + <span class="hljs-params">1</span> <span class="hljs-keyword">else</span> dp[<span class="hljs-params">-1</span>] ``` ``` **復雜度分析** - 時間復雜度:O(amonut?len(coins))O(amonut \* len(coins))O(amonut?len(coins)) - 空間復雜度:O(amount)O(amount)O(amount) ## 擴展 這是一道很簡單描述的題目, 因此很多時候會被用到大公司的電面中。 相似問題: [518.coin-change-2](https://github.com/azl397985856/leetcode/blob/master/problems/518.coin-change-2.md)
                  <ruby id="bdb3f"></ruby>

                  <p id="bdb3f"><cite id="bdb3f"></cite></p>

                    <p id="bdb3f"><cite id="bdb3f"><th id="bdb3f"></th></cite></p><p id="bdb3f"></p>
                      <p id="bdb3f"><cite id="bdb3f"></cite></p>

                        <pre id="bdb3f"></pre>
                        <pre id="bdb3f"><del id="bdb3f"><thead id="bdb3f"></thead></del></pre>

                        <ruby id="bdb3f"><mark id="bdb3f"></mark></ruby><ruby id="bdb3f"></ruby>
                        <pre id="bdb3f"><pre id="bdb3f"><mark id="bdb3f"></mark></pre></pre><output id="bdb3f"></output><p id="bdb3f"></p><p id="bdb3f"></p>

                        <pre id="bdb3f"><del id="bdb3f"><progress id="bdb3f"></progress></del></pre>

                              <ruby id="bdb3f"></ruby>

                              哎呀哎呀视频在线观看