{ "type": "code_options", "author": "https://github.com/qiyuangong/leetcode", "source": "solution.md", "exercise_id": "af25ce9a4f4c425b8c900b3345aef661", "keywords": "数组,双指针", "title": "移除元素", "desc": [ { "content": "\n
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给你一个数组 nums 和一个值 val,你需要 原地\n移除所有数值等于 val 的元素,并返回移除后数组的新长度。

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不要使用额外的数组空间,你必须仅使用 O(1) 额外空间并 原地 修改输入数组

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元素的顺序可以改变。你不需要考虑数组中超出新长度后面的元素。

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说明:

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为什么返回数值是整数,但输出的答案是数组呢?

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请注意,输入数组是以「引用」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。

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你可以想象内部操作如下:

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// nums 是以“引用”方式传递的。也就是说,不对实参作任何拷贝\nint len = removeElement(nums, val);",
      "language": "markdown"
    },
    {
      "content": "\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 该长度范围内 的所有元素。\nfor (int i = 0; i < len; i++) {\n    print(nums[i]);\n}\n
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示例 1:

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输入:nums = [3,2,2,3], val = 3\n
输出:
2, nums = [2,2]\n
解释:
函数应该返回新的长度 2, 并且 nums 中的前两个元素均为 2。你不需要考虑数组中超出新长度后面的元素。例如,函数返回的新长度为 2 ,而 nums = [2,2,3,3] 或 nums = [2,2,0,0],也会被视作正确答案。\n
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示例 2:

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输入:nums = [0,1,2,2,3,0,4,2], val = 2\n
输出:
5, nums = [0,1,4,0,3]\n
解释:
函数应该返回新的长度 5, 并且 nums 中的前五个元素为 0, 1, 3, 0, 4。注意这五个元素可为任意顺序。你不需要考虑数组中超出新长度后面的元素。\n
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提示:

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", "language": "markdown" } ], "answer": [ { "content": "", "language": "python" } ], "prepared": [ [ { "content": "", "language": "python" } ], [ { "content": "", "language": "python" } ], [ { "content": "", "language": "python" } ] ], "template": { "content": "class Solution(object):\n\tdef removeElement(self, nums, val):\n\t\tls = len(nums)\n\t\tif ls == 0:\n\t\t\treturn ls\n\t\tcount = 0\n\t\tindex = 0\n\t\twhile index < ls - count:\n\t\t\tif nums[index] == val:\n\t\t\t\tnums[index] = nums[ls - 1 - count]\n\t\t\t\tcount += 1\n\t\t\telse:\n\t\t\t\tindex += 1\n\t\treturn ls - count\nif __name__ == '__main__':\n\ts = Solution()\n\tprint(s.removeElement([3,2,2,3], 3))", "language": "python" }, "node_id": "dailycode-f46ab61031ab4af6926cc6db067d1b78", "license": "csdn.net", "created_at": 1637894161, "topic_link": "https://bbs.csdn.net/topics/600470219" }