{ "data": { "question": { "questionId": "2285", "questionFrontendId": "2166", "categoryTitle": "Algorithms", "boundTopicId": 1243456, "title": "Design Bitset", "titleSlug": "design-bitset", "content": "
A Bitset is a data structure that compactly stores bits.
\n\nImplement the Bitset class:
Bitset(int size) Initializes the Bitset with size bits, all of which are 0.void fix(int idx) Updates the value of the bit at the index idx to 1. If the value was already 1, no change occurs.void unfix(int idx) Updates the value of the bit at the index idx to 0. If the value was already 0, no change occurs.void flip() Flips the values of each bit in the Bitset. In other words, all bits with value 0 will now have value 1 and vice versa.boolean all() Checks if the value of each bit in the Bitset is 1. Returns true if it satisfies the condition, false otherwise.boolean one() Checks if there is at least one bit in the Bitset with value 1. Returns true if it satisfies the condition, false otherwise.int count() Returns the total number of bits in the Bitset which have value 1.String toString() Returns the current composition of the Bitset. Note that in the resultant string, the character at the ith index should coincide with the value at the ith bit of the Bitset.\n
Example 1:
\n\n\nInput\n["Bitset", "fix", "fix", "flip", "all", "unfix", "flip", "one", "unfix", "count", "toString"]\n[[5], [3], [1], [], [], [0], [], [], [0], [], []]\nOutput\n[null, null, null, null, false, null, null, true, null, 2, "01010"]\n\nExplanation\nBitset bs = new Bitset(5); // bitset = "00000".\nbs.fix(3); // the value at idx = 3 is updated to 1, so bitset = "00010".\nbs.fix(1); // the value at idx = 1 is updated to 1, so bitset = "01010". \nbs.flip(); // the value of each bit is flipped, so bitset = "10101". \nbs.all(); // return False, as not all values of the bitset are 1.\nbs.unfix(0); // the value at idx = 0 is updated to 0, so bitset = "00101".\nbs.flip(); // the value of each bit is flipped, so bitset = "11010". \nbs.one(); // return True, as there is at least 1 index with value 1.\nbs.unfix(0); // the value at idx = 0 is updated to 0, so bitset = "01010".\nbs.count(); // return 2, as there are 2 bits with value 1.\nbs.toString(); // return "01010", which is the composition of bitset.\n\n\n
\n
Constraints:
\n\n1 <= size <= 1050 <= idx <= size - 1105 calls will be made in total to fix, unfix, flip, all, one, count, and toString.all, one, count, or toString.5 calls will be made to toString.位集 Bitset 是一种能以紧凑形式存储位的数据结构。
\n\n请你实现 Bitset 类。
Bitset(int size) 用 size 个位初始化 Bitset ,所有位都是 0 。void fix(int idx) 将下标为 idx 的位上的值更新为 1 。如果值已经是 1 ,则不会发生任何改变。void unfix(int idx) 将下标为 idx 的位上的值更新为 0 。如果值已经是 0 ,则不会发生任何改变。void flip() 翻转 Bitset 中每一位上的值。换句话说,所有值为 0 的位将会变成 1 ,反之亦然。boolean all() 检查 Bitset 中 每一位 的值是否都是 1 。如果满足此条件,返回 true ;否则,返回 false 。boolean one() 检查 Bitset 中 是否 至少一位 的值是 1 。如果满足此条件,返回 true ;否则,返回 false 。int count() 返回 Bitset 中值为 1 的位的 总数 。String toString() 返回 Bitset 的当前组成情况。注意,在结果字符串中,第 i 个下标处的字符应该与 Bitset 中的第 i 位一致。\n\n
示例:
\n\n\n输入\n[\"Bitset\", \"fix\", \"fix\", \"flip\", \"all\", \"unfix\", \"flip\", \"one\", \"unfix\", \"count\", \"toString\"]\n[[5], [3], [1], [], [], [0], [], [], [0], [], []]\n输出\n[null, null, null, null, false, null, null, true, null, 2, \"01010\"]\n\n解释\nBitset bs = new Bitset(5); // bitset = \"00000\".\nbs.fix(3); // 将 idx = 3 处的值更新为 1 ,此时 bitset = \"00010\" 。\nbs.fix(1); // 将 idx = 1 处的值更新为 1 ,此时 bitset = \"01010\" 。\nbs.flip(); // 翻转每一位上的值,此时 bitset = \"10101\" 。\nbs.all(); // 返回 False ,bitset 中的值不全为 1 。\nbs.unfix(0); // 将 idx = 0 处的值更新为 0 ,此时 bitset = \"00101\" 。\nbs.flip(); // 翻转每一位上的值,此时 bitset = \"11010\" 。\nbs.one(); // 返回 True ,至少存在一位的值为 1 。\nbs.unfix(0); // 将 idx = 0 处的值更新为 0 ,此时 bitset = \"01010\" 。\nbs.count(); // 返回 2 ,当前有 2 位的值为 1 。\nbs.toString(); // 返回 \"01010\" ,即 bitset 的当前组成情况。\n\n\n
\n\n
提示:
\n\n1 <= size <= 1050 <= idx <= size - 1fix、unfix、flip、all、one、count 和 toString 方法 总共 105 次all、one、count 或 toString 方法一次toString 方法 5 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>, 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