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<li class="md-nav__item">
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<a href="../../chapter_preface/installation/" class="md-nav__link">
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0.3. 编程环境安装
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0.4. 一起参与创作
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<label class="md-nav__link" for="__nav_3" id="__nav_3_label" tabindex="0">
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2. 计算复杂度
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2. 复杂度分析
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2. 计算复杂度
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2. 复杂度分析
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12. 附录
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12. 附录
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<a href="../../chapter_appendix/installation/" class="md-nav__link">
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12.1. 编程环境安装
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</a>
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<a href="../../chapter_appendix/contribution/" class="md-nav__link">
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12.2. 一起参与创作
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</a>
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参考文献
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</label>
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<p>你可以假设每种输入只会对应一个答案。但是,数组中同一个元素在答案里不能重复出现。</p>
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<p>你可以按任意顺序返回答案。</p>
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</div>
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<p>「暴力枚举」和「辅助哈希表」分别为 <strong>空间最优</strong> 和 <strong>时间最优</strong> 的两种解法。本着时间比空间更宝贵的原则,后者是本题的最佳解法。</p>
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<p>「暴力枚举」和「辅助哈希表」分别对应 <strong>空间最优</strong> 和 <strong>时间最优</strong> 的两种解法。本着时间比空间更宝贵的原则,后者是本题的最佳解法。</p>
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<h3 id="_1">方法一:暴力枚举<a class="headerlink" href="#_1" title="Permanent link">¶</a></h3>
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<p>时间复杂度 <span class="arithmatex">\(O(N^2)\)</span> ,空间复杂度 <span class="arithmatex">\(O(1)\)</span> ,属于「时间换空间」。</p>
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<p>虽然仅使用常数大小的额外空间,但运行速度过慢。</p>
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<p>考虑直接遍历所有所有可能性。通过开启一个两层循环,判断两个整数的和是否为 <code>target</code> ,若是则返回它俩的索引(即下标)即可。</p>
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<div class="tabbed-set tabbed-alternate" data-tabs="1:10"><input checked="checked" id="__tabbed_1_1" name="__tabbed_1" type="radio" /><input id="__tabbed_1_2" name="__tabbed_1" type="radio" /><input id="__tabbed_1_3" name="__tabbed_1" type="radio" /><input id="__tabbed_1_4" name="__tabbed_1" type="radio" /><input id="__tabbed_1_5" name="__tabbed_1" type="radio" /><input id="__tabbed_1_6" name="__tabbed_1" type="radio" /><input id="__tabbed_1_7" name="__tabbed_1" type="radio" /><input id="__tabbed_1_8" name="__tabbed_1" type="radio" /><input id="__tabbed_1_9" name="__tabbed_1" type="radio" /><input id="__tabbed_1_10" name="__tabbed_1" type="radio" /><div class="tabbed-labels"><label for="__tabbed_1_1">Java</label><label for="__tabbed_1_2">C++</label><label for="__tabbed_1_3">Python</label><label for="__tabbed_1_4">Go</label><label for="__tabbed_1_5">JavaScript</label><label for="__tabbed_1_6">TypeScript</label><label for="__tabbed_1_7">C</label><label for="__tabbed_1_8">C#</label><label for="__tabbed_1_9">Swift</label><label for="__tabbed_1_10">Zig</label></div>
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<p>该方法的时间复杂度为 <span class="arithmatex">\(O(N^2)\)</span> ,空间复杂度为 <span class="arithmatex">\(O(1)\)</span> ,<strong>属于时间换空间</strong>。本方法时间复杂度较高,在大数据量下非常耗时。</p>
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<h3 id="_2">方法二:辅助哈希表<a class="headerlink" href="#_2" title="Permanent link">¶</a></h3>
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<p>时间复杂度 <span class="arithmatex">\(O(N)\)</span> ,空间复杂度 <span class="arithmatex">\(O(N)\)</span> ,属于「空间换时间」。</p>
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<p>借助辅助哈希表 dic ,通过保存数组元素与索引的映射来提升算法运行速度。</p>
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<p>考虑借助一个哈希表,key 为数组元素、value 为元素索引。循环遍历数组中的每个元素 <code>num</code> ,并执行:</p>
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<ol>
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<li>判断数字 <code>target - num</code> 是否在哈希表中,若是则直接返回该两个元素的索引;</li>
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<li>将元素 <code>num</code> 和其索引添加进哈希表;</li>
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</ol>
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<div class="tabbed-set tabbed-alternate" data-tabs="2:10"><input checked="checked" id="__tabbed_2_1" name="__tabbed_2" type="radio" /><input id="__tabbed_2_2" name="__tabbed_2" type="radio" /><input id="__tabbed_2_3" name="__tabbed_2" type="radio" /><input id="__tabbed_2_4" name="__tabbed_2" type="radio" /><input id="__tabbed_2_5" name="__tabbed_2" type="radio" /><input id="__tabbed_2_6" name="__tabbed_2" type="radio" /><input id="__tabbed_2_7" name="__tabbed_2" type="radio" /><input id="__tabbed_2_8" name="__tabbed_2" type="radio" /><input id="__tabbed_2_9" name="__tabbed_2" type="radio" /><input id="__tabbed_2_10" name="__tabbed_2" type="radio" /><div class="tabbed-labels"><label for="__tabbed_2_1">Java</label><label for="__tabbed_2_2">C++</label><label for="__tabbed_2_3">Python</label><label for="__tabbed_2_4">Go</label><label for="__tabbed_2_5">JavaScript</label><label for="__tabbed_2_6">TypeScript</label><label for="__tabbed_2_7">C</label><label for="__tabbed_2_8">C#</label><label for="__tabbed_2_9">Swift</label><label for="__tabbed_2_10">Zig</label></div>
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<p>该方法的时间复杂度为 <span class="arithmatex">\(O(N)\)</span> ,空间复杂度为 <span class="arithmatex">\(O(N)\)</span> ,<strong>体现空间换时间</strong>。本方法虽然引入了额外空间使用,但时间和空间使用整体更加均衡,因此为本题最优解法。</p>
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