剑指 Offer 59 - II:队列的最大值

src/main/java/leetcode/editor/cn/DuiLieDeZuiDaZhiLcof.java

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+//请定义一个队列并实现函数 max_value 得到队列里的最大值，要求函数max_value、push_back 和 pop_front 的均摊时间复杂度都
+//是O(1)。 
+//
+// 若队列为空，pop_front 和 max_value 需要返回 -1 
+//
+// 示例 1： 
+//
+// 输入: 
+//["MaxQueue","push_back","push_back","max_value","pop_front","max_value"]
+//[[],[1],[2],[],[],[]]
+//输出: [null,null,null,2,1,2]
+// 
+//
+// 示例 2： 
+//
+// 输入: 
+//["MaxQueue","pop_front","max_value"]
+//[[],[],[]]
+//输出: [null,-1,-1]
+// 
+//
+// 
+//
+// 限制： 
+//
+// 
+// 1 <= push_back,pop_front,max_value的总操作数 <= 10000 
+// 1 <= value <= 10^5 
+// 
+// Related Topics 栈 Sliding Window 
+// 👍 248 👎 0
+
+package leetcode.editor.cn;
+
+//剑指 Offer 59 - II:队列的最大值
+public class DuiLieDeZuiDaZhiLcof {
+    public static void main(String[] args) {
+        //测试代码
+//        Solution solution = new DuiLieDeZuiDaZhiLcof().new Solution();
+    }
+
+    //力扣代码
+    //leetcode submit region begin(Prohibit modification and deletion)
+    class MaxQueue {
+        int[] arr = new int[20000];
+        int start = 0, end = 0;
+
+        public MaxQueue() {
+
+        }
+
+        public int max_value() {
+            int ans = -1;
+            for (int i = start; i != end; ++i) {
+                ans = Math.max(ans, arr[i]);
+            }
+            return ans;
+        }
+
+        public void push_back(int value) {
+            arr[end++] = value;
+        }
+
+        public int pop_front() {
+            if (start == end) {
+                return -1;
+            }
+            return arr[start++];
+        }
+    }
+
+/**
+ * Your MaxQueue object will be instantiated and called as such:
+ * MaxQueue obj = new MaxQueue();
+ * int param_1 = obj.max_value();
+ * obj.push_back(value);
+ * int param_3 = obj.pop_front();
+ */
+//leetcode submit region end(Prohibit modification and deletion)
+
+}

src/main/java/leetcode/editor/cn/DuiLieDeZuiDaZhiLcof.md

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+<p>请定义一个队列并实现函数 <code>max_value</code> 得到队列里的最大值，要求函数<code>max_value</code>、<code>push_back</code> 和 <code>pop_front</code> 的<strong>均摊</strong>时间复杂度都是O(1)。</p>
+
+<p>若队列为空，<code>pop_front</code> 和 <code>max_value</code>&nbsp;需要返回 -1</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre><strong>输入:</strong> 
+[&quot;MaxQueue&quot;,&quot;push_back&quot;,&quot;push_back&quot;,&quot;max_value&quot;,&quot;pop_front&quot;,&quot;max_value&quot;]
+[[],[1],[2],[],[],[]]
+<strong>输出:&nbsp;</strong>[null,null,null,2,1,2]
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre><strong>输入:</strong> 
+[&quot;MaxQueue&quot;,&quot;pop_front&quot;,&quot;max_value&quot;]
+[[],[],[]]
+<strong>输出:&nbsp;</strong>[null,-1,-1]
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>限制：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= push_back,pop_front,max_value的总操作数&nbsp;&lt;= 10000</code></li>
+	<li><code>1 &lt;= value &lt;= 10^5</code></li>
+</ul>
+<div><div>Related Topics</div><div><li>栈</li><li>Sliding Window</li></div></div>\n<div><li>👍 248</li><li>👎 0</li></div>