<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>

                合規國際互聯網加速 OSASE為企業客戶提供高速穩定SD-WAN國際加速解決方案。 廣告
                # 0337. 打家劫舍 III ## 題目地址(337. 打家劫舍 III) <https://leetcode-cn.com/problems/house-robber-iii/> ## 題目描述 ``` <pre class="calibre18">``` 在上次打劫完一條街道之后和一圈房屋后,小偷又發現了一個新的可行竊的地區。這個地區只有一個入口,我們稱之為“根”。 除了“根”之外,每棟房子有且只有一個“父“房子與之相連。一番偵察之后,聰明的小偷意識到“這個地方的所有房屋的排列類似于一棵二叉樹”。 如果兩個直接相連的房子在同一天晚上被打劫,房屋將自動報警。 計算在不觸動警報的情況下,小偷一晚能夠盜取的最高金額。 示例 1: 輸入: [3,2,3,null,3,null,1] 3 / \ 2 3 \ \ 3 1 輸出: 7 解釋: 小偷一晚能夠盜取的最高金額 = 3 + 3 + 1 = 7. 示例 2: 輸入: [3,4,5,1,3,null,1] 3 / \ 4 5 / \ \ 1 3 1 輸出: 9 解釋: 小偷一晚能夠盜取的最高金額 = 4 + 5 = 9. ``` ``` ## 前置知識 - 二叉樹 - 動態規劃 ## 公司 - 阿里 - 騰訊 - 百度 - 字節 ## 思路 和 198.house-robber 類似,這道題也是相同的思路。 只不過數據結構從數組換成了樹。 我們仍然是對每一項進行決策:**如果我搶的話,所得到的最大價值是多少。如果我不搶的話,所得到的最大價值是多少。** - 遍歷二叉樹,都每一個節點我們都需要判斷搶還是不搶。 - 如果搶了的話, 那么我們不能繼續搶其左右子節點 - 如果不搶的話,那么我們可以繼續搶左右子節點,當然也可以不搶。搶不搶取決于哪個價值更大。 - 搶不搶取決于哪個價值更大。 這是一個明顯的遞歸問題,我們使用遞歸來解決。由于沒有重復子問題,因此沒有必要 cache ,也沒有必要動態規劃。 ## 關鍵點 - 對每一個節點都分析,是搶還是不搶 ## 代碼 語言支持:JS, C++,Java,Python JavaScript Code: ``` <pre class="calibre18">``` <span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">helper</span>(<span class="hljs-params">root</span>) </span>{ <span class="hljs-keyword">if</span> (root === <span class="hljs-params">null</span>) <span class="hljs-keyword">return</span> [<span class="hljs-params">0</span>, <span class="hljs-params">0</span>]; <span class="hljs-title">// 0: rob 1: notRob</span> <span class="hljs-keyword">const</span> l = helper(root.left); <span class="hljs-keyword">const</span> r = helper(root.right); <span class="hljs-keyword">const</span> robed = root.val + l[<span class="hljs-params">1</span>] + r[<span class="hljs-params">1</span>]; <span class="hljs-keyword">const</span> notRobed = <span class="hljs-params">Math</span>.max(l[<span class="hljs-params">0</span>], l[<span class="hljs-params">1</span>]) + <span class="hljs-params">Math</span>.max(r[<span class="hljs-params">0</span>], r[<span class="hljs-params">1</span>]); <span class="hljs-keyword">return</span> [robed, notRobed]; } <span class="hljs-title">/** * @param {TreeNode} root * @return {number} */</span> <span class="hljs-keyword">var</span> rob = <span class="hljs-function"><span class="hljs-keyword">function</span> (<span class="hljs-params">root</span>) </span>{ <span class="hljs-keyword">const</span> [robed, notRobed] = helper(root); <span class="hljs-keyword">return</span> <span class="hljs-params">Math</span>.max(robed, notRobed); }; ``` ``` C++ Code: ``` <pre class="calibre18">``` <span class="hljs-title">/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */</span> <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">rob</span><span class="hljs-params">(TreeNode* root)</span> </span>{ pair<<span class="hljs-keyword">int</span>, <span class="hljs-keyword">int</span>> res = dfs(root); <span class="hljs-keyword">return</span> max(res.first, res.second); } pair<<span class="hljs-keyword">int</span>, <span class="hljs-keyword">int</span>> dfs(TreeNode* root) { pair<<span class="hljs-keyword">int</span>, <span class="hljs-keyword">int</span>> res = {<span class="hljs-params">0</span>, <span class="hljs-params">0</span>}; <span class="hljs-keyword">if</span>(root == <span class="hljs-params">NULL</span>) { <span class="hljs-keyword">return</span> res; } pair<<span class="hljs-keyword">int</span>, <span class="hljs-keyword">int</span>> left = dfs(root->left); pair<<span class="hljs-keyword">int</span>, <span class="hljs-keyword">int</span>> right = dfs(root->right); <span class="hljs-title">// 0 代表不偷,1 代表偷</span> res.first = max(left.first, left.second) + max(right.first, right.second); res.second = left.first + right.first + root->val; <span class="hljs-keyword">return</span> res; } }; ``` ``` Java Code: ``` <pre class="calibre18">``` <span class="hljs-title">/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */</span> <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">public</span> <span class="hljs-keyword">int</span> <span class="hljs-title">rob</span><span class="hljs-params">(TreeNode root)</span> </span>{ <span class="hljs-keyword">int</span>[] res = dfs(root); <span class="hljs-keyword">return</span> Math.max(res[<span class="hljs-params">0</span>], res[<span class="hljs-params">1</span>]); } <span class="hljs-keyword">public</span> <span class="hljs-keyword">int</span>[] dp(TreeNode root) { <span class="hljs-keyword">int</span>[] res = <span class="hljs-keyword">new</span> <span class="hljs-keyword">int</span>[<span class="hljs-params">2</span>]; <span class="hljs-keyword">if</span>(root == <span class="hljs-keyword">null</span>) { <span class="hljs-keyword">return</span> res; } <span class="hljs-keyword">int</span>[] left = dfs(root.left); <span class="hljs-keyword">int</span>[] right = dfs(root.right); <span class="hljs-title">// 0 代表不偷,1 代表偷</span> res[<span class="hljs-params">0</span>] = Math.max(left[<span class="hljs-params">0</span>], left[<span class="hljs-params">1</span>]) + Math.max(right[<span class="hljs-params">0</span>], right[<span class="hljs-params">1</span>]); res[<span class="hljs-params">1</span>] = left[<span class="hljs-params">0</span>] + right[<span class="hljs-params">0</span>] + root.val; <span class="hljs-keyword">return</span> res; } } ``` ``` Python 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">rob</span><span class="hljs-params">(self, root: TreeNode)</span> -> int:</span> <span class="hljs-function"><span class="hljs-keyword">def</span> <span class="hljs-title">dfs</span><span class="hljs-params">(node)</span>:</span> <span class="hljs-keyword">if</span> <span class="hljs-keyword">not</span> node: <span class="hljs-keyword">return</span> [<span class="hljs-params">0</span>, <span class="hljs-params">0</span>] [l_rob, l_not_rob] = dfs(node.left) [r_rob, r_not_rob] = dfs(node.right) <span class="hljs-keyword">return</span> [node.val + l_not_rob + r_not_rob, max([l_rob, l_not_rob]) + max([r_rob, r_not_rob])] <span class="hljs-keyword">return</span> max(dfs(root)) <span class="hljs-title"># @lc code=end</span> ``` ``` **復雜度分析** - 時間復雜度:O(N)O(N)O(N),其中 N 為樹的節點個數。 - 空間復雜度:O(h)O(h)O(h),其中 h 為樹的高度。 ## 相關題目 - [198.house-robber](https://github.com/azl397985856/leetcode/blob/master/problems/198.house-robber.md) 大家對此有何看法,歡迎給我留言,我有時間都會一一查看回答。更多算法套路可以訪問我的 LeetCode 題解倉庫:<https://github.com/azl397985856/leetcode> 。 目前已經 37K star 啦。 大家也可以關注我的公眾號《力扣加加》帶你啃下算法這塊硬骨頭。 ![](https://img.kancloud.cn/a3/63/a363818092b0356fbd67882f0389528b_900x500.jpg)
                  <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