{ "question_id": 26, "question_title": "删除有序数组中的重复项", "difficulty": "简单", "question_content": "<div class=\"notranslate\">\n <p>给你一个有序数组 <code>nums</code> ，请你<strong><a href=\"http://baike.baidu.com/item/%E5%8E%9F%E5%9C%B0%E7%AE%97%E6%B3%95\">\n 原地</a></strong> 删除重复出现的元素，使每个元素 <strong>只出现一次</strong> ，返回删除后数组的新长度。</p>\n\n <p>不要使用额外的数组空间，你必须在 <strong><a href=\"https://baike.baidu.com/item/%E5%8E%9F%E5%9C%B0%E7%AE%97%E6%B3%95\">原地\n </a>修改输入数组 </strong>并在使用 O(1) 额外空间的条件下完成。</p>\n\n <p> </p>\n\n <p><strong>说明:</strong></p>\n\n <p>为什么返回数值是整数，但输出的答案是数组呢?</p>\n\n <p>请注意，输入数组是以<strong>「引用」</strong>方式传递的，这意味着在函数里修改输入数组对于调用者是可见的。</p>\n\n <p>你可以想象内部操作如下:</p>\n\n <pre>// <strong>nums</strong> 是以“引用”方式传递的。也就是说，不对实参做任何拷贝\nint len = removeDuplicates(nums);\n\n// 在函数里修改输入数组对于调用者是可见的。\n// 根据你的函数返回的长度, 它会打印出数组中<strong> 该长度范围内</strong> 的所有元素。\nfor (int i = 0; i < len; i++) {\n print(nums[i]);\n}\n</pre>\n \n\n <p><strong>示例 1：</strong></p>\n\n <pre><strong>输入：</strong>nums = [1,1,2]\n<strong><br />输出：</strong>2, nums = [1,2]\n<strong><br />解释：</strong>函数应该返回新的长度 <strong>2</strong> ，并且原数组 <em>nums </em>的前两个元素被修改为 <strong>1</strong>, <strong>2 </strong>。不需要考虑数组中超出新长度后面的元素。\n </pre>\n\n <p><strong>示例 2：</strong></p>\n\n <pre><strong>输入：</strong>nums = [0,0,1,1,1,2,2,3,3,4]\n<strong><br />输出：</strong>5, nums = [0,1,2,3,4]\n<strong><br />解释：</strong>函数应该返回新的长度 <strong>5</strong> ， 并且原数组 <em>nums </em>的前五个元素被修改为 <strong>0</strong>, <strong>1</strong>, <strong>2</strong>, <strong>3</strong>, <strong>4</strong> 。不需要考虑数组中超出新长度后面的元素。\n </pre>\n\n <p> </p>\n\n <p><strong>提示：</strong></p>\n\n <ul>\n <li><code>0 <= nums.length <= 3 * 10<sup>4</sup></code></li>\n <li><code>-10<sup>4</sup> <= nums[i] <= 10<sup>4</sup></code></li>\n <li><code>nums</code> 已按升序排列</li>\n </ul>\n\n <p> </p>\n</div>", "topic_link": "https://bbs.csdn.net/topics/600470118", "cpp": "#include <bits/stdc++.h>\nusing namespace std;\nclass Solution\n{\npublic:\n\tint removeDuplicates(vector<int> &nums)\n\t{\n\t\tif (nums.size() == 0)\n\t\t{\n\t\t\treturn 0;\n\t\t}\n\t\tint count = 1;\n\t\tfor (int i = 1; i < nums.size(); i++)\n\t\t{\n\t\t\tif (nums[i - 1] != nums[i])\n\t\t\t{\n\t\t\t\tnums[count++] = nums[i];\n\t\t\t}\n\t\t}\n\t\treturn count;\n\t}\n};", "java": "class Solution {\n\tpublic int removeDuplicates(int[] nums) {\n\t\tif (nums == null || nums.length == 0) {\n\t\t\treturn 0;\n\t\t}\n\t\tint a = 0;\n\t\tfor (int b = 1; b < nums.length; b++) {\n\t\t\tif (nums[a] != nums[b]) {\n\t\t\t\ta++;\n\t\t\t\tnums[a] = nums[b];\n\t\t\t}\n\t\t}\n\t\treturn a + 1;\n\t}\n}\n", "js": "", "python": "class Solution(object):\n\tdef removeDuplicates(self, nums):\n\t\tif len(nums) == 0:\n\t\t\treturn 0\n\t\tleft = 0\n\t\tfor i in range(1, len(nums)):\n\t\t\tif nums[left] == nums[i]:\n\t\t\t\tcontinue\n\t\t\telse:\n\t\t\t\tleft += 1\n\t\t\t\tnums[left] = nums[i]\n\t\treturn left + 1\n# %%\ns = Solution()\nprint(s.removeDuplicates(nums = [1,1,2]))", "status": 1, "keywords": "数组,双指针", "license": { "cpp": "https://github.com/begeekmyfriend/leetcode", "python": "https://github.com/qiyuangong/leetcode", "java": "https://github.com/zhangyu345293721/leetcode" }, "notebook": { "cpp": "https://codechina.csdn.net/csdn/csdn-daily-code/-/jupyter/master/data/notebook/leetcode/ipynb/25/25_cpp.ipynb?type=file", "python": "https://codechina.csdn.net/csdn/csdn-daily-code/-/jupyter/master/data/notebook/leetcode/ipynb/25/25_python.ipynb?type=file", "java": "https://codechina.csdn.net/csdn/csdn-daily-code/-/jupyter/master/data/notebook/leetcode/ipynb/25/25_java.ipynb?type=file" }, "notebook_enable": 1 }
