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柳婼の blog – 我不管，反正我最萌~

LeetCode 461. Hamming Distance. The Hamming distance between two integers is the number of positions at which the corresponding bits are different. Given two integers x and y, calculate the Hamming distance. Input: x = 1, y = 4. 1 (0 0 0 1). 4 (0 1 0 0). The above arrows point to positions where the corresponding bits are different. 对于右移一位，采用x / 2和 y / 2的方式，对于比较最后一位，即比较x % 2 和 y % 2，统计不相同的次数cnt，直到x和y都等于0为止. Class Solution { public: int hammingDistance(int x, int y) { int cnt = 0; while(x! 0) { if(x % 2!

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柳婼の blog – 我不管，反正我最萌~ | liuchuo.net Reviews

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LeetCode 461. Hamming Distance. The Hamming distance between two integers is the number of positions at which the corresponding bits are different. Given two integers x and y, calculate the Hamming distance. Input: x = 1, y = 4. 1 (0 0 0 1). 4 (0 1 0 0). The above arrows point to positions where the corresponding bits are different. 对于右移一位，采用x / 2和 y / 2的方式，对于比较最后一位，即比较x % 2 和 y % 2，统计不相同的次数cnt，直到x和y都等于0为止. Class Solution { public: int hammingDistance(int x, int y) { int cnt = 0; while(x! 0) { if(x % 2!

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柳婼の blog – 我不管，反正我最萌~ | liuchuo.net Reviews

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LeetCode 461. Hamming Distance. The Hamming distance between two integers is the number of positions at which the corresponding bits are different. Given two integers x and y, calculate the Hamming distance. Input: x = 1, y = 4. 1 (0 0 0 1). 4 (0 1 0 0). The above arrows point to positions where the corresponding bits are different. 对于右移一位，采用x / 2和 y / 2的方式，对于比较最后一位，即比较x % 2 和 y % 2，统计不相同的次数cnt，直到x和y都等于0为止. Class Solution { public: int hammingDistance(int x, int y) { int cnt = 0; while(x! 0) { if(x % 2!

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天梯赛GPLT题解目录 – 柳婼の blog

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Github源码地址 https:/ www.github.com/liuchuo/PAT.

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GPLT – 柳婼の blog

http://www.liuchuo.net/archives/tag/gplt

X is the root x是根结点. X and y are siblings x和y是兄弟结点. X is the parent of y x是y的父结点. X is a child of y x是y的一个子结点。每组测试第1行包含2个正整数N = 1000 和M = 20 ，分别是插入元素的个数、以及需要判断的命题数。对输入的每个命题，如果其为真，则在一行中输出 T ，否则输出 F。 46 23 26 24 10. 24 is the root. 26 and 23 are siblings. 46 is the parent of 23. 23 is a child of 10. Include cstdio #include vector #include cstring using namespace std; vector int v; int n; void upAdjust(int i) { if(i = 1) return ; while(i! 输入第一行给出一个正整N = 1000 第2行给出N个正整数，以空格分隔。 88 74 101 26 15 0 34 22 77.

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PAT乙级题解目录 – 柳婼の blog

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源码地址 https:/ github.com/liuchuo/PAT.

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LeetCode题解目录 – 柳婼の blog

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源码地址 https:/ github.com/liuchuo/LeetCode. Longest Substring Without Repeating Characters. Median of Two Sorted Arrays. String to Integer (atoi). Container With Most Water. Letter Combinations of a Phone Number. Remove Nth Node From End of List. Merge Two Sorted Lists. Merge k Sorted Lists. Swap Nodes in Pairs. Reverse Nodes in k-Group. Remove Duplicates from Sorted Array. Substring with Concatenation of All Words. Search in Rotated Sorted Array. Search for a Range. Length of Last Word. Search a 2D Matrix.

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LeetCode OJ – 柳婼の blog

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LeetCode 524. Longest Word in Dictionary through Deleting. Given a string and a string dictionary, find the longest string in the dictionary that can be formed by deleting some characters of the given string. If there are more than one possible results, return the longest word with the smallest lexicographical order. If there is no possible result, return the empty string. S = “abpcplea”, d = [“ale”,”apple”,”monkey”,”plea”]. S = “abpcplea”, d = [“a”,”b”,”c”]. Note: There are at least two nodes in this BST.

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博客报错 A Parser-blocking, cross-origin script, is invoked via document.write. A Parser-blocking, cross-origin script, https:/ s95.cnzz.com/z stat.php? Id=1257684520, is invoked via document.write. This may be blocked by the browser if the device has poor network connectivity. A Parser-blocking, cross-origin script, https:/ c.cnzz.com/core.php? Web id=1257684520&t=z, is invoked via document.write. This may be blocked by the browser if the device has poor network connectivity. 在验证管理添加验证后获取的 id 和 key。用到的技术细节...

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柳婼の blog – 我不管，反正我最萌~

LeetCode 461. Hamming Distance. The Hamming distance between two integers is the number of positions at which the corresponding bits are different. Given two integers x and y, calculate the Hamming distance. Input: x = 1, y = 4. 1 (0 0 0 1). 4 (0 1 0 0). The above arrows point to positions where the corresponding bits are different. 对于右移一位，采用x / 2和 y / 2的方式，对于比较最后一位，即比较x % 2 和 y % 2，统计不相同的次数cnt，直到x和y都等于0为止. Class Solution { public: int hammingDistance(int x, int y) { int cnt = 0; while(x! 0) { if(x % 2!

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