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<p>整数数组 <code>nums</code> 按升序排列,数组中的值 <strong>互不相同</strong> 。</p>
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<p>在传递给函数之前,<code>nums</code> 在预先未知的某个下标 <code>k</code>(<code>0 <= k < nums.length</code>)上进行了 <strong>旋转</strong>,使数组变为 <code>[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]</code>(下标 <strong>从 0 开始</strong> 计数)。例如, <code>[0,1,2,4,5,6,7]</code> 在下标 <code>3</code> 处经旋转后可能变为 <code>[4,5,6,7,0,1,2]</code> 。</p>
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<p>在传递给函数之前,<code>nums</code> 在预先未知的某个下标 <code>k</code>(<code>0 <= k < nums.length</code>)上进行了 <strong>向左旋转</strong>,使数组变为 <code>[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]</code>(下标 <strong>从 0 开始</strong> 计数)。例如, <code>[0,1,2,4,5,6,7]</code> 下标 <code>3</code> 上向左旋转后可能变为 <code>[4,5,6,7,0,1,2]</code> 。</p>
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<p>给你 <strong>旋转后</strong> 的数组 <code>nums</code> 和一个整数 <code>target</code> ,如果 <code>nums</code> 中存在这个目标值 <code>target</code> ,则返回它的下标,否则返回 <code>-1</code> 。</p>
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<p>给你一个只包含 <code>'('</code> 和 <code>')'</code> 的字符串,找出最长有效(格式正确且连续)括号<span data-keyword="substring">子串</span>的长度。</p>
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<p>给你一个只包含 <code>'('</code> 和 <code>')'</code> 的字符串,找出最长有效(格式正确且连续)括号 <span data-keyword="substring">子串</span> 的长度。</p>
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<p>左右括号匹配,即每个左括号都有对应的右括号将其闭合的字符串是格式正确的,比如 <code>"(()())"</code>。</p>
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<p> </p>
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<p><span class="example-io"><b>输出:</b>true</span></p>
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</div>
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<p><strong class="example">示例 5:</strong></p>
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<div class="example-block">
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<p><span class="example-io"><b>输入:</b>s = "([)]"</span></p>
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<p><span class="example-io"><b>输出:</b>false</span></p>
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</div>
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<p> </p>
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<p><strong>提示:</strong></p>
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