leetCode191/201/202/136 -Number of 1 Bits/Bitwise AND of Numbers Range/Happy Number/Single Number

一：Number of 1 Bits

题目：

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.

Credits:

Special thanks to @ts for adding this problem and creating all test cases.

解法一：

此题关键是怎样推断一个数字的第i为是否为0  即： x& (1<<i)

class Solution {
public:
    int hammingWeight(uint32_t n) {
        int count = 0;
        for(int i = 0; i < 32; i++){
            if((n & (1<<i)) != 0)count++;
        }
        return count;

    }
};

解法二：此解关键在于明确n&(n-1)会n最后一位1消除，这样循环下去就能够求出n的位数中为1的个数

class Solution {
public:
    int hammingWeight(uint32_t n) {
        int count = 0;
        while(n > 0){
            n &= n-1;
            count ++;
        }
        return count;
    }
};

二：Bitwise AND
of Numbers Range

题目：

Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers in this range, inclusive.

For example, given the range [5, 7], you should return 4.

分析：此题提供两种解法：1：当m到n之前假设跨过了1,2,4,8等2^i次方的数字时(即推断m与n是否具有同样的最高位)，则会为0，否则顺序将m到n相与。

解法二：利用上题中的思路。n&(n-1)会消除n中最后一个1，如1100000100当与n-1按位与时便会消除最后一个1，赋值给n(这样就减免了非常多不必要按位与的过程)

解法一：

class Solution {
public:
    int rangeBitwiseAnd(int m, int n) {
        int bitm = 0, bitn = 0;
        for(int i =0; i < 31; i++){
            if(m & (1<<i))bitm = i;
            if(n & (1<<i))bitn = i;
        }
        if(bitm == bitn){
            int sum = m;
            for(int i = m; i < n; i++)  // 为了防止 2147483647+1 超过范围
                sum = (sum & i);
            sum = (sum & n);
            return sum;
        }
        else return 0;
    }
};

解法二：

class Solution {
public:
    int rangeBitwiseAnd(int m, int n) {
        while(n > m){
            n &= n-1;
        }
        return n;
    }
};

三：Happy Number

题目：

Write an algorithm to determine if a number is "happy".

A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle
which does not include 1. Those numbers for which this process ends in 1 are happy numbers.

Example: 19 is a happy number

• 12 + 92 = 82
• 82 + 22 = 68
• 62 + 82 = 100
• 12 + 02 + 02 =
  1

分析：此题关键是用一个set或者map来存储该数字是否已经出现过————hash_map+math

class Solution {
public:
    bool isHappy(int n) {
        while(n != 1){
            if(hset.count(n)) return false;    // 通过hashtable 推断是否出现过
            hset.insert(n);
            int sum = 0;
            while(n != 0){    // 求元素的各个位置平方和
                int mod = n%10;
                n = n/10;
                sum += mod * mod;
            }
            n = sum;
        }
        return true;

    }
private:
    set<int> hset;
};

四：Single Number

题目：

Given an array of integers, every element appears twice except for one. Find that single one.

Note:

Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

分析：此题关键在于用到异或

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int ans = 0;
        for(int i = 0; i < nums.size(); i++)
            ans ^= nums[i];
        return ans;

    }
};