leetcode-problemset/leetcode-cn/originData/maximum-frequency-stack.json

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        "question": {
            "questionId": "931",
            "questionFrontendId": "895",
            "categoryTitle": "Algorithms",
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            "title": "Maximum Frequency Stack",
            "titleSlug": "maximum-frequency-stack",
            "content": "<p>Design a stack-like data structure to push elements to the stack and pop the most frequent element from the stack.</p>\n\n<p>Implement the <code>FreqStack</code> class:</p>\n\n<ul>\n\t<li><code>FreqStack()</code> constructs an empty frequency stack.</li>\n\t<li><code>void push(int val)</code> pushes an integer <code>val</code> onto the top of the stack.</li>\n\t<li><code>int pop()</code> removes and returns the most frequent element in the stack.\n\t<ul>\n\t\t<li>If there is a tie for the most frequent element, the element closest to the stack&#39;s top is removed and returned.</li>\n\t</ul>\n\t</li>\n</ul>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input</strong>\n[&quot;FreqStack&quot;, &quot;push&quot;, &quot;push&quot;, &quot;push&quot;, &quot;push&quot;, &quot;push&quot;, &quot;push&quot;, &quot;pop&quot;, &quot;pop&quot;, &quot;pop&quot;, &quot;pop&quot;]\n[[], [5], [7], [5], [7], [4], [5], [], [], [], []]\n<strong>Output</strong>\n[null, null, null, null, null, null, null, 5, 7, 5, 4]\n\n<strong>Explanation</strong>\nFreqStack freqStack = new FreqStack();\nfreqStack.push(5); // The stack is [5]\nfreqStack.push(7); // The stack is [5,7]\nfreqStack.push(5); // The stack is [5,7,5]\nfreqStack.push(7); // The stack is [5,7,5,7]\nfreqStack.push(4); // The stack is [5,7,5,7,4]\nfreqStack.push(5); // The stack is [5,7,5,7,4,5]\nfreqStack.pop();   // return 5, as 5 is the most frequent. The stack becomes [5,7,5,7,4].\nfreqStack.pop();   // return 7, as 5 and 7 is the most frequent, but 7 is closest to the top. The stack becomes [5,7,5,4].\nfreqStack.pop();   // return 5, as 5 is the most frequent. The stack becomes [5,7,4].\nfreqStack.pop();   // return 4, as 4, 5 and 7 is the most frequent, but 4 is closest to the top. The stack becomes [5,7].\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>0 &lt;= val &lt;= 10<sup>9</sup></code></li>\n\t<li>At most <code>2 * 10<sup>4</sup></code> calls will be made to <code>push</code> and <code>pop</code>.</li>\n\t<li>It is guaranteed that there will be at least one element in the stack before calling <code>pop</code>.</li>\n</ul>\n",
            "translatedTitle": "最大频率栈",
            "translatedContent": "<p>设计一个类似堆栈的数据结构，将元素推入堆栈，并从堆栈中弹出<strong>出现频率</strong>最高的元素。</p>\n\n<p>实现 <code>FreqStack</code>&nbsp;类:</p>\n\n<ul>\n\t<li><meta charset=\"UTF-8\" /><code>FreqStack()</code>&nbsp;构造一个空的堆栈。</li>\n\t<li><meta charset=\"UTF-8\" /><code>void push(int val)</code>&nbsp;将一个整数&nbsp;<code>val</code>&nbsp;压入栈顶。</li>\n\t<li><meta charset=\"UTF-8\" /><code>int pop()</code>&nbsp;删除并返回堆栈中出现频率最高的元素。\n\t<ul>\n\t\t<li>如果出现频率最高的元素不只一个，则移除并返回最接近栈顶的元素。</li>\n\t</ul>\n\t</li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>\n[\"FreqStack\",\"push\",\"push\",\"push\",\"push\",\"push\",\"push\",\"pop\",\"pop\",\"pop\",\"pop\"],\n[[],[5],[7],[5],[7],[4],[5],[],[],[],[]]\n<strong>输出：</strong>[null,null,null,null,null,null,null,5,7,5,4]\n<strong>解释：</strong>\nFreqStack = new FreqStack();\nfreqStack.push (5);//堆栈为 [5]\nfreqStack.push (7);//堆栈是 [5,7]\nfreqStack.push (5);//堆栈是 [5,7,5]\nfreqStack.push (7);//堆栈是 [5,7,5,7]\nfreqStack.push (4);//堆栈是 [5,7,5,7,4]\nfreqStack.push (5);//堆栈是 [5,7,5,7,4,5]\nfreqStack.pop ();//返回 5 ，因为 5 出现频率最高。堆栈变成 [5,7,5,7,4]。\nfreqStack.pop ();//返回 7 ，因为 5 和 7 出现频率最高，但7最接近顶部。堆栈变成 [5,7,5,4]。\nfreqStack.pop ();//返回 5 ，因为 5 出现频率最高。堆栈变成 [5,7,4]。\nfreqStack.pop ();//返回 4 ，因为 4, 5 和 7 出现频率最高，但 4 是最接近顶部的。堆栈变成 [5,7]。</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>0 &lt;= val &lt;= 10<sup>9</sup></code></li>\n\t<li><code>push</code>&nbsp;和 <code>pop</code>&nbsp;的操作数不大于 <code>2 * 10<sup>4</sup></code>。</li>\n\t<li>输入保证在调用&nbsp;<code>pop</code>&nbsp;之前堆栈中至少有一个元素。</li>\n</ul>\n",
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                    "code": "class FreqStack {\npublic:\n    FreqStack() {\n\n    }\n    \n    void push(int val) {\n\n    }\n    \n    int pop() {\n\n    }\n};\n\n/**\n * Your FreqStack object will be instantiated and called as such:\n * FreqStack* obj = new FreqStack();\n * obj->push(val);\n * int param_2 = obj->pop();\n */",
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                    "code": "public class FreqStack {\n\n    public FreqStack() {\n\n    }\n    \n    public void Push(int val) {\n\n    }\n    \n    public int Pop() {\n\n    }\n}\n\n/**\n * Your FreqStack object will be instantiated and called as such:\n * FreqStack obj = new FreqStack();\n * obj.Push(val);\n * int param_2 = obj.Pop();\n */",
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                    "code": "type FreqStack struct {\n\n}\n\n\nfunc Constructor() FreqStack {\n\n}\n\n\nfunc (this *FreqStack) Push(val int)  {\n\n}\n\n\nfunc (this *FreqStack) Pop() int {\n\n}\n\n\n/**\n * Your FreqStack object will be instantiated and called as such:\n * obj := Constructor();\n * obj.Push(val);\n * param_2 := obj.Pop();\n */",
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                    "code": "class FreqStack\n    def initialize()\n\n    end\n\n\n=begin\n    :type val: Integer\n    :rtype: Void\n=end\n    def push(val)\n\n    end\n\n\n=begin\n    :rtype: Integer\n=end\n    def pop()\n\n    end\n\n\nend\n\n# Your FreqStack object will be instantiated and called as such:\n# obj = FreqStack.new()\n# obj.push(val)\n# param_2 = obj.pop()",
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