LeetCode:Reverse Bits Reverse bits of a given 32 bits unsigned integer. For example, given input 43261596 (represented in binary as00000010100101000001111010011100), return 964176192 (represented in binary as00111001011110000010100101000000). Follow up: If this function is called many times, how would you...
1classSolution {2public:3uint32_t reverseBits(uint32_t n) {4uint32_t answer;5uint32_tinti;67answer =0;89/*把一个unsigned int 数字1一直左移,直到它变成全0的时候,也就得到了该机器内unsigned int的长度*/10for(i =1; i !=0; i <<=1)11{12answer <<=1;13if(value &1) { answer |...
1uint32_t reverseBits(uint32_t n) {//uint32_t位4个字节(4*8=32),int根据操作系统不同,字节数不同,16位操作为2字节,32位操作系统为4字节,64位为8字节 2uint32_t ret=0;3for(inti=0; i<32; i++) {//为什么没有考虑将十进制转换为二进制?? 因为用到& | << >>操作后,计算机就默认将整...
直接代码: 1classSolution:2#@param n, an integer3#@return an integer4defreverseBits(self, n):5tmp=bin(n)6tmp=tmp[2:]7iflen(tmp)!=32:8forainrange(32-len(tmp)):9tmp='0'+tmp10tmp=tmp[::-1]11returnint(tmp,2)
LeetCode算法题-Reverse Bits(Java实现) 这是悦乐书的第185次更新,第187篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第44题(顺位题号是190)。给定32位无符号整数,求它的反转位。例如: 输入:43261596 输出:964176192 说明:43261596以二进制表示为00000010100101000001111010011100,...
下边的程序中用到了bitset数据结构,需要注意的是,bits[0]才是最低位。 1classSolution {2public:3uint32_t reverseBits(uint32_t n) {4uint32_t one =1;5bitset<32>bits;6inti =31;7while(n !=0&& i > -1)8{9bits[i] = n &one;10n = n >>1;11i--;12}1314returnbits.to_ulong();15...
LeetCode 190. Reverse Bits (反转位) 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)....
题目地址:https://leetcode.com/problems/reverse-bits/description/题目描述Reverse bits of a given 32 bits unsigned integer.Example:Input: 43261596 Output: 964176192 Explanation: 43261596 represented in binary as 00000010100101000001111010011100, return 964176192 represented in binary as ...
classSolution{public:uint32_treverseBits(uint32_tn){uint32_tres =0;for(inti=0; i<32; i++) { res = (res <<1) | (n &0x1); n>>=1; }returnres; } }; 古典解法 class Solution { public: uint32_t reverseBits(uint32_t n) { ...
Reverse Integer那道题会考虑溢出,因为那是reverse each digit,这里不会溢出,因为reverse的是each bit 1publicclassSolution {2//you need treat n as an unsigned value3publicintreverseBits(intn) {4intres = 0;5for(inti=0; i<32; i++) {6intbit = (n>>i) & 1;7res |= bit<<(31-i);8}...