publicinthammingWeight(intn){intcount =0;for(inti=0; i<32; i++){if((n&1) ==1) { count++; } n = n >>1; }returncount; } 03 第二种解法 对于上面的解法,还可以再简化下,省掉那个判断,因为和1做与运算,如果最后一位是1,整个运算结果就是1,就相当于是自增加1。 publicinthammingWeight...
【LeetCode191】Number of 1 Bits★ 1.题目 2.思路 方法一:常规方法。 方法二:给面试官惊喜的解法。 3.java代码 方法一代码: 1publicclassSolution {2//you need to treat n as an unsigned value3publicinthammingWeight(intn) {4intcount=0;5intflag=1;6while(flag!=0){7if((n&flag)!=0)8count...
For example, the 32-bit integer ’11’ has binary representation 00000000000000000000000000001011, so the function should return 3. 就是检查1的个数,和上一道题leetcode 190. Reverse Bits的思想有点类似,值得学习。 代码如下: public class Solution { public int hammingWeight(int n) { int count = 0...
Write a function that takes an unsigned integer and return the number of '1' bits it has (also known as the Hamming weight). Example 1: Input: 00000000000000000000000000001011 Output: 3 Explanation: The input binary string 00000000000000000000000000001011 has a total of three '1' bits. Example 2...
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1. Description Sort Integers by The Number of 1 Bits 2. Solution 解析:由于最大数字不超过10000,因此1的位数不超过14位,注意+=的运算优先级要低于&,而+的运算优先级要高于&。Version 1用右移运算获得1的个数,Version 2用的bin函数,Version 3通过python自带的排序函数进行排序。Version 4使用字典保存结果。
Given an integer array arr. You have to sort the integers in the array in ascending order by the number of1'sin their binary representation and in case of two or more integers have the same number of1'syou have to sort them in ascending order. ...
var countPrimeSetBits = function(L, R) { // 存储二进制中1的个数 let arr = [] for (let i = L; i <= R; i++) { arr.push(i.toString(2).replace(/0/g, '').length) } // 判断一个数是否为质数 let isPrime = function(num) { if (num === 1) { re...
1866. 恰有 K 根木棍可以看到的排列数目 - 有 n 根长度互不相同的木棍,长度为从 1 到 n 的整数。请你将这些木棍排成一排,并满足从左侧 可以看到 恰好 k 根木棍。从左侧 可以看到 木棍的前提是这个木棍的 左侧 不存在比它 更长的 木棍。 * 例如,如果木棍排列为 [1,3,2,5
Write a function that takes an unsigned integer and returns the number of ’1' bits it has (also known as the Hamming weight). For example, the 32-bit integer ’11' has binary representation 00000000000000000000000000001011, so the function should return 3. ...