You are given an integer n representing the size of a 0-indexed memory array. All memory units are initially free.
You have a memory allocator with the following functionalities:
\n\nsize consecutive free memory units and assign it the id mID.mID.Note that:
\n\nmID.mID, even if they were allocated in different blocks.Implement the Allocator class:
Allocator(int n) Initializes an Allocator object with a memory array of size n.int allocate(int size, int mID) Find the leftmost block of size consecutive free memory units and allocate it with the id mID. Return the block's first index. If such a block does not exist, return -1.int free(int mID) Free all memory units with the id mID. Return the number of memory units you have freed.\n
Example 1:
\n\n\nInput\n["Allocator", "allocate", "allocate", "allocate", "free", "allocate", "allocate", "allocate", "free", "allocate", "free"]\n[[10], [1, 1], [1, 2], [1, 3], [2], [3, 4], [1, 1], [1, 1], [1], [10, 2], [7]]\nOutput\n[null, 0, 1, 2, 1, 3, 1, 6, 3, -1, 0]\n\nExplanation\nAllocator loc = new Allocator(10); // Initialize a memory array of size 10. All memory units are initially free.\nloc.allocate(1, 1); // The leftmost block's first index is 0. The memory array becomes [1,_,_,_,_,_,_,_,_,_]. We return 0.\nloc.allocate(1, 2); // The leftmost block's first index is 1. The memory array becomes [1,2,_,_,_,_,_,_,_,_]. We return 1.\nloc.allocate(1, 3); // The leftmost block's first index is 2. The memory array becomes [1,2,3,_,_,_,_,_,_,_]. We return 2.\nloc.free(2); // Free all memory units with mID 2. The memory array becomes [1,_, 3,_,_,_,_,_,_,_]. We return 1 since there is only 1 unit with mID 2.\nloc.allocate(3, 4); // The leftmost block's first index is 3. The memory array becomes [1,_,3,4,4,4,_,_,_,_]. We return 3.\nloc.allocate(1, 1); // The leftmost block's first index is 1. The memory array becomes [1,1,3,4,4,4,_,_,_,_]. We return 1.\nloc.allocate(1, 1); // The leftmost block's first index is 6. The memory array becomes [1,1,3,4,4,4,1,_,_,_]. We return 6.\nloc.free(1); // Free all memory units with mID 1. The memory array becomes [_,_,3,4,4,4,_,_,_,_]. We return 3 since there are 3 units with mID 1.\nloc.allocate(10, 2); // We can not find any free block with 10 consecutive free memory units, so we return -1.\nloc.free(7); // Free all memory units with mID 7. The memory array remains the same since there is no memory unit with mID 7. We return 0.\n\n\n
\n
Constraints:
\n\n1 <= n, size, mID <= 10001000 calls will be made to allocate and free.给你一个整数 n ，表示下标从 0 开始的内存数组的大小。所有内存单元开始都是空闲的。
请你设计一个具备以下功能的内存分配器：
\n\nsize 的连续空闲内存单元并赋 id mID 。mID 对应的所有内存单元。注意：
\n\nmID 。mID 对应的所有内存单元，即便这些内存单元被分配在不同的块中。实现 Allocator 类：
Allocator(int n) 使用一个大小为 n 的内存数组初始化 Allocator 对象。int allocate(int size, int mID) 找出大小为 size 个连续空闲内存单元且位于 最左侧 的块，分配并赋 id mID 。返回块的第一个下标。如果不存在这样的块，返回 -1 。int free(int mID) 释放 id mID 对应的所有内存单元。返回释放的内存单元数目。\n\n
示例：
\n\n输入\n[\"Allocator\", \"allocate\", \"allocate\", \"allocate\", \"free\", \"allocate\", \"allocate\", \"allocate\", \"free\", \"allocate\", \"free\"]\n[[10], [1, 1], [1, 2], [1, 3], [2], [3, 4], [1, 1], [1, 1], [1], [10, 2], [7]]\n输出\n[null, 0, 1, 2, 1, 3, 1, 6, 3, -1, 0]\n\n解释\nAllocator loc = new Allocator(10); // 初始化一个大小为 10 的内存数组，所有内存单元都是空闲的。\nloc.allocate(1, 1); // 最左侧的块的第一个下标是 0 。内存数组变为 [1, , , , , , , , , ]。返回 0 。\nloc.allocate(1, 2); // 最左侧的块的第一个下标是 1 。内存数组变为 [1,2, , , , , , , , ]。返回 1 。\nloc.allocate(1, 3); // 最左侧的块的第一个下标是 2 。内存数组变为 [1,2,3, , , , , , , ]。返回 2 。\nloc.free(2); // 释放 mID 为 2 的所有内存单元。内存数组变为 [1, ,3, , , , , , , ] 。返回 1 ，因为只有 1 个 mID 为 2 的内存单元。\nloc.allocate(3, 4); // 最左侧的块的第一个下标是 3 。内存数组变为 [1, ,3,4,4,4, , , , ]。返回 3 。\nloc.allocate(1, 1); // 最左侧的块的第一个下标是 1 。内存数组变为 [1,1,3,4,4,4, , , , ]。返回 1 。\nloc.allocate(1, 1); // 最左侧的块的第一个下标是 6 。内存数组变为 [1,1,3,4,4,4,1, , , ]。返回 6 。\nloc.free(1); // 释放 mID 为 1 的所有内存单元。内存数组变为 [ , ,3,4,4,4, , , , ] 。返回 3 ，因为有 3 个 mID 为 1 的内存单元。\nloc.allocate(10, 2); // 无法找出长度为 10 个连续空闲内存单元的空闲块，所有返回 -1 。\nloc.free(7); // 释放 mID 为 7 的所有内存单元。内存数组保持原状，因为不存在 mID 为 7 的内存单元。返回 0 。\n\n\n
\n\n
提示：
\n\n1 <= n, size, mID <= 1000allocate 和 free 方法 1000 次\\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\"],\"java\":[\"Java\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u88ab\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5305\\u542b Pair \\u7c7b: https:\\/\\/docs.oracle.com\\/javase\\/8\\/javafx\\/api\\/javafx\\/util\\/Pair.html <\\/p>\"],\"python\":[\"Python\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982\\uff1aarray<\\/a>, bisect<\\/a>, collections<\\/a>\\u3002\\u5982\\u679c\\u60a8\\u9700\\u8981\\u4f7f\\u7528\\u5176\\u4ed6\\u5e93\\u51fd\\u6570\\uff0c\\u8bf7\\u81ea\\u884c\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u6ce8\\u610f Python 2.7 \\u5c06\\u57282020\\u5e74\\u540e\\u4e0d\\u518d\\u7ef4\\u62a4<\\/a>\\u3002 \\u5982\\u60f3\\u4f7f\\u7528\\u6700\\u65b0\\u7248\\u7684Python\\uff0c\\u8bf7\\u9009\\u62e9Python 3\\u3002<\\/p>\"],\"c\":[\"C\",\" \\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5982\\u60f3\\u4f7f\\u7528\\u54c8\\u5e0c\\u8868\\u8fd0\\u7b97, \\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 uthash<\\/a>\\u3002 \\\"uthash.h\\\"\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5bfc\\u5165\\u3002\\u8bf7\\u770b\\u5982\\u4e0b\\u793a\\u4f8b:<\\/p>\\r\\n\\r\\n 1. 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\\u6a21\\u5757\\u4e2d\\u63d0\\u4f9b\\uff1ahttps:\\/\\/www.rubydoc.info\\/github\\/kanwei\\/algorithms\\/Algorithms<\\/p>\"],\"swift\":[\"Swift\",\" \\u7248\\u672c\\uff1a \\u6211\\u4eec\\u901a\\u5e38\\u4fdd\\u8bc1\\u66f4\\u65b0\\u5230 Apple\\u653e\\u51fa\\u7684\\u6700\\u65b0\\u7248Swift<\\/a>\\u3002\\u5982\\u679c\\u60a8\\u53d1\\u73b0Swift\\u4e0d\\u662f\\u6700\\u65b0\\u7248\\u7684\\uff0c\\u8bf7\\u8054\\u7cfb\\u6211\\u4eec\\uff01\\u6211\\u4eec\\u5c06\\u5c3d\\u5feb\\u66f4\\u65b0\\u3002<\\/p>\"],\"golang\":[\"Go\",\" \\u7248\\u672c\\uff1a \\u652f\\u6301 https:\\/\\/godoc.org\\/github.com\\/emirpasic\\/gods@v1.18.1<\\/a> \\u7b2c\\u4e09\\u65b9\\u5e93\\u3002<\\/p>\"],\"python3\":[\"Python3\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982array<\\/a>, bisect<\\/a>, collections<\\/a>\\u3002 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AddressSanitizer<\\/a>\\u4e5f\\u88ab\\u5f00\\u542f\\u6765\\u68c0\\u6d4bout-of-bounds<\\/code>\\u548cuse-after-free<\\/code>\\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\n\\r\\nstruct hash_entry {\\r\\n int id; \\/* we'll use this field as the key *\\/\\r\\n char name[10];\\r\\n UT_hash_handle hh; \\/* makes this structure hashable *\\/\\r\\n};\\r\\n\\r\\nstruct hash_entry *users = NULL;\\r\\n\\r\\nvoid add_user(struct hash_entry *s) {\\r\\n HASH_ADD_INT(users, id, s);\\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\\r\\n\\r\\n\\r\\nstruct hash_entry *find_user(int user_id) {\\r\\n struct hash_entry *s;\\r\\n HASH_FIND_INT(users, &user_id, s);\\r\\n return s;\\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\\r\\n\\r\\n\\r\\nvoid delete_user(struct hash_entry *user) {\\r\\n HASH_DEL(users, user); \\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\"],\"csharp\":[\"C#\",\"Node.js 16.13.2<\\/code><\\/p>\\r\\n\\r\\n--harmony<\\/code> \\u6807\\u8bb0\\u6765\\u5f00\\u542f \\u65b0\\u7248ES6\\u7279\\u6027<\\/a>\\u3002<\\/p>\\r\\n\\r\\nRuby 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