题目: Reverse bits of a given 32 bits unsigned integer. For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 00111001011110000010100101000000). Follow up: If this func…

题目 Reverse bits of a given 32 bits unsigned integer. For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 00111001011110000010100101000000). Follow up: If this funct…

题目: Reverse digits of an integer. Example1: x = 123, return 321 Example2: x = -123, return -321 分析: 乍看,好似是一个很简单的题目,只需要将整数从最低位起到最高位依次处理即可,但是,此题的关键在于如何处理溢出数据.我们知道,Integer类型数据的范围是: #define       INT_MIN         (-2147483647 - 1)          /* minimum (sig…

翻转32位无符号二进制整数 Reverse bits of a given 32 bits unsigned integer. For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 00111001011110000010100101000000). Follow up:If…

题目 Given an input string, reverse the string word by word. For example, Given s = "the sky is blue", return "blue is sky the". Update (2015-02-12): For C programmers: Try to solve it in-place in O(1) space. click to show clarification.…

题目 Reverse a singly linked list. click to show more hints. Hint: A linked list can be reversed either iteratively or recursively. Could you implement both? 分析 反转链表. 一个简单的解法,既然反转该链表,我们把所有节点作为一个输入序列,按照头插法重新构造一个链表即可,它既是所给链表的反转结果. 题目所给提示还有另外两种方法解决,迭代和递归.…

题目 Reverse a linked list from position m to n. Do it in-place and in one-pass. For example: Given 1->2->3->4->5->NULL, m = 2 and n = 4, return 1->4->3->2->5->NULL. Note: Given m, n satisfy the following condition: 1 ≤ m ≤ n ≤…

题目 Given a linked list, reverse the nodes of a linked list k at a time and return its modified list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is. You may not alter the values in the nodes, only…

Medium! 题目描述: 给定一个可包含重复数字的序列,返回所有不重复的全排列. 示例: 输入: [1,1,2] 输出: [ [1,1,2], [1,2,1], [2,1,1] ] 解题思路: 这道题是之前那道Permutations 全排列的延伸,由于输入数组有可能出现重复数字,如果按照之前的算法运算,会有重复排列产生,我们要避免重复的产生,在递归函数中要判断前面一个数和当前的数是否相等,如果相等,前面的数必须已经使用了,即对应的visited中的值为1,当前的数字才能使用,否则需要跳过,这…

题目 Given a collection of numbers that might contain duplicates, return all possible unique permutations. For example, [1,1,2] have the following unique permutations: [1,1,2], [1,2,1], and [2,1,1]. 分析 用上一题的代码,完全可以AC,那是因为我们的库函数next_permutation()以及prev_…