A bit flip of a number x is choosing a bit in the binary representation of x and flipping it from either 0 to 1 or 1 to 0.
x = 7, the binary representation is 111 and we may choose any bit (including any leading zeros not shown) and flip it. We can flip the first bit from the right to get 110, flip the second bit from the right to get 101, flip the fifth bit from the right (a leading zero) to get 10111, etc.Given two integers start and goal, return the minimum number of bit flips to convert start to goal.
\n
Example 1:
\n\n\nInput: start = 10, goal = 7\nOutput: 3\nExplanation: The binary representation of 10 and 7 are 1010 and 0111 respectively. We can convert 10 to 7 in 3 steps:\n- Flip the first bit from the right: 1010 -> 1011.\n- Flip the third bit from the right: 1011 -> 1111.\n- Flip the fourth bit from the right: 1111 -> 0111.\nIt can be shown we cannot convert 10 to 7 in less than 3 steps. Hence, we return 3.\n\n
Example 2:
\n\n\nInput: start = 3, goal = 4\nOutput: 3\nExplanation: The binary representation of 3 and 4 are 011 and 100 respectively. We can convert 3 to 4 in 3 steps:\n- Flip the first bit from the right: 011 -> 010.\n- Flip the second bit from the right: 010 -> 000.\n- Flip the third bit from the right: 000 -> 100.\nIt can be shown we cannot convert 3 to 4 in less than 3 steps. Hence, we return 3.\n\n\n
\n
Constraints:
\n\n0 <= start, goal <= 109\n
Note: This question is the same as 461: Hamming Distance.
\n", "translatedTitle": "转换数字的最少位翻转次数", "translatedContent": "一次 位翻转 定义为将数字 x 二进制中的一个位进行 翻转 操作，即将 0 变成 1 ，或者将 1 变成 0 。
x = 7 ，二进制表示为 111 ，我们可以选择任意一个位（包含没有显示的前导 0 ）并进行翻转。比方说我们可以翻转最右边一位得到 110 ，或者翻转右边起第二位得到 101 ，或者翻转右边起第五位（这一位是前导 0 ）得到 10111 等等。给你两个整数 start 和 goal ，请你返回将 start 转变成 goal 的 最少位翻转 次数。
\n\n
示例 1：
\n\n\n输入：start = 10, goal = 7\n输出：3\n解释：10 和 7 的二进制表示分别为 1010 和 0111 。我们可以通过 3 步将 10 转变成 7 ：\n- 翻转右边起第一位得到：1010 -> 1011 。\n- 翻转右边起第三位：1011 -> 1111 。\n- 翻转右边起第四位：1111 -> 0111 。\n我们无法在 3 步内将 10 转变成 7 。所以我们返回 3 。\n\n
示例 2：
\n\n\n输入：start = 3, goal = 4\n输出：3\n解释：3 和 4 的二进制表示分别为 011 和 100 。我们可以通过 3 步将 3 转变成 4 ：\n- 翻转右边起第一位：011 -> 010 。\n- 翻转右边起第二位：010 -> 000 。\n- 翻转右边起第三位：000 -> 100 。\n我们无法在 3 步内将 3 变成 4 。所以我们返回 3 。\n\n\n
\n\n
提示：
\n\n0 <= start, goal <= 109\n\n
注意：本题与 461. 汉明距离 相同。
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}clang 19<\\/code> \\u91c7\\u7528\\u6700\\u65b0 C++ 23 \\u6807\\u51c6\\uff0c\\u5e76\\u4f7f\\u7528 GCC 14 \\u63d0\\u4f9b\\u7684 libstdc++<\\/code>\\u3002<\\/p>\\r\\n\\r\\n-O2<\\/code> \\u7ea7\\u4f18\\u5316\\uff0c\\u5e76\\u63d0\\u4f9b -gline-tables-only<\\/code> \\u53c2\\u6570\\u3002AddressSanitizer<\\/a> \\u4e5f\\u88ab\\u5f00\\u542f\\u6765\\u68c0\\u6d4b out-of-bounds<\\/code> \\u548c use-after-free<\\/code> \\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\nOpenJDK 21<\\/code>\\u3002\\u4f7f\\u7528\\u7f16\\u8bd1\\u53c2\\u6570 --enable-preview --release 21<\\/code><\\/p>\\r\\n\\r\\nPython 2.7.18<\\/code><\\/p>\\r\\n\\r\\nGCC 14<\\/code>\\uff0c\\u91c7\\u7528 GNU11 \\u6807\\u51c6\\u3002<\\/p>\\r\\n\\r\\n-O2<\\/code> \\u7ea7\\u4f18\\u5316\\uff0c\\u5e76\\u63d0\\u4f9b -g1<\\/code> \\u53c2\\u6570\\u3002 AddressSanitizer<\\/a> \\u4e5f\\u88ab\\u5f00\\u542f\\u6765\\u68c0\\u6d4b out-of-bounds<\\/code> \\u548c use-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 20.10.0<\\/code><\\/p>\\r\\n\\r\\n--harmony<\\/code> \\u6807\\u8bb0\\u6765\\u5f00\\u542f \\u65b0\\u7248ES6\\u7279\\u6027<\\/a>\\u3002<\\/p>\\r\\n\\r\\nRuby 3.2<\\/code> \\u6267\\u884c<\\/p>\\r\\n\\r\\nSwift 6.0<\\/code><\\/p>\\r\\n\\r\\nGo 1.23<\\/code><\\/p>\\r\\n\\r\\nPython 3.11<\\/code><\\/p>\\r\\n\\r\\nScala 3.3.1<\\/code><\\/p>\"],\"kotlin\":[\"Kotlin\",\"Kotlin 1.9.0<\\/code><\\/p>\\r\\n\\r\\nrust 1.79.0<\\/code><\\/p>\\r\\n\\r\\nPHP 8.2<\\/code>.<\\/p>\\r\\n\\r\\n-O2 --disable-reflection<\\/code><\\/p>\\r\\n\\r\\n