<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>

                ??碼云GVP開源項目 12k star Uniapp+ElementUI 功能強大 支持多語言、二開方便! 廣告
                ## [面試題64. 求1+2+…+n](https://leetcode-cn.com/problems/qiu-12n-lcof/) >Easy #### 思路 一眼看到題目就蹦出一句話,`首項加尾項乘以項數除以2` -- 等差數組求和公式。 題目要求不能使用乘除、條件語句,于是優先想到兩種方案: * 內置函數 * 位運算 挑個簡單的,AC! 時間復雜度 `$ O(n) $` 空間復雜度 `$ O(n) $` #### 代碼 python3 ``` class Solution: def sumNums(self, n: int) -> int: return sum([x for x in range(1,n+1)]) ```
                  <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