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LeetCode 53. Maximum Subarray.html
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<meta property="og:description" content="53. Maximum Subarray（最大子序和）链接https://leetcode-cn.com/problems/maximum-subarray 题目给定一个整数数组 nums ，找到一个具有最大和的连续子数组（子数组最少包含一个元素），返回其最大和。 示例: 输入: [-2,1,-3,4,-1,2,1,-5,4], 输出: 6 解释: 连续子数组 [4,-1,2,1] 的和最">
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<h2 id="53-Maximum-Subarray（最大子序和）"><a href="#53-Maximum-Subarray（最大子序和）" class="headerlink" title="53. Maximum Subarray（最大子序和）"></a>53. Maximum Subarray（最大子序和）</h2><h3 id="链接"><a href="#链接" class="headerlink" title="链接"></a>链接</h3><p><a href="https://leetcode-cn.com/problems/maximum-subarray" target="_blank" rel="noopener">https://leetcode-cn.com/problems/maximum-subarray</a></p>
<h3 id="题目"><a href="#题目" class="headerlink" title="题目"></a>题目</h3><p>给定一个整数数组 nums ，找到一个具有最大和的连续子数组（子数组最少包含一个元素），返回其最大和。</p>
<p>示例:</p>
<p> 输入: [-2,1,-3,4,-1,2,1,-5,4],<br> 输出: 6<br> 解释: 连续子数组 [4,-1,2,1] 的和最大，为 6。<br>进阶:</p>
<p> 如果你已经实现复杂度为 O(n) 的解法，尝试使用更为精妙的分治法求解。</p>
<h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><p>没尝试分治法，那个有点麻烦，等做到别的分治时候再做。<br>这边有动态规划的思想在里面，从头到尾遍历，如果和小于0，那么直接无视，重新开始，和大于0，就加上之前的和。每次都保留当前最大的和，直到输出。</p>
<h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br></pre></td><td class="code"><pre><span class="line">public static int maxSubArray(int[] nums) {</span><br><span class="line"> int ans = nums[0];</span><br><span class="line"> int sum = 0;</span><br><span class="line"> for (int num : nums) {</span><br><span class="line"> if (sum >= 0) {</span><br><span class="line"> sum += num;</span><br><span class="line"> } else {</span><br><span class="line"> sum = num;</span><br><span class="line"> }</span><br><span class="line"> ans = Math.max(sum, ans);</span><br><span class="line"> }</span><br><span class="line"> return ans;</span><br><span class="line"> }</span><br></pre></td></tr></table></figure>
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