Top 150 · 位运算（6 题）

class Solution {
    public String addBinary(String a, String b) {
        StringBuilder sb = new StringBuilder();
        if(a.length() < b.length()){
            String temp = b;
            b = a;
            a = temp;
        }
        
        int i = a.length() - 1;
        int j = b.length() - 1;
        
        boolean carry =false;
        while(j >=0){
            if(a.charAt(i) =='0' && b.charAt(j) == '0'){
                if(carry){
                    sb.append("1");
                    carry = false;
                }else{
                    sb.append("0");
                }
            }else if(a.charAt(i) =='1' && b.charAt(j) == '1'){
                if(carry){
                    sb.append("1");
                }else{
                    sb.append("0");
                    carry = true;
                }
            }else{
                if(carry){
                    sb.append("0");
                    carry = true;
                }else{
                    sb.append("1");
                }
            }
            j--;
            i--;
        }
        
        while (i >=0) {
            if(a.charAt(i) =='0'){
                if(carry){
                    sb.append("1");
                    carry = false;
                }else{
                    sb.append("0");
                }
            }else{
                if(carry){
                    sb.append("0");
                }else{
                    sb.append("1");
                }
            }
            i--;
            j--;
        }
        if(carry){
            sb.append("1");
        }
        return sb.reverse().toString();
    }
}

class Solution {
    private static final int m0 = 0x55555555; // 01010101 ...
    private static final int m1 = 0x33333333; // 00110011 ...
    private static final int m2 = 0x0f0f0f0f; // 00001111 ...
    private static final int m3 = 0x00ff00ff; // 00000000111111110000000011111111

    public int reverseBits(int n) {
        n = n>>>1&m0 | (n&m0)<<1; // 交换相邻位
        n = n>>>2&m1 | (n&m1)<<2; // 两个两个交换
        n = n>>>4&m2 | (n&m2)<<4; // 四个四个交换
        n = n>>>8&m3 | (n&m3)<<8; // 八个八个交换
        return n>>>16 | n<<16;    // 交换高低 16 位
    }
}

class Solution {
    public int reverseBits(int n) {
        return Integer.reverse(n);
    }
}

class Solution {
    public int hammingWeight(int n) {
        int count =0;
        
        while(n>0){
            if(n%2==1){
                 count +=1;
            }
            n >>= 1;
        }
        return count;
    }
}

class Solution {
    public int singleNumber(int[] nums) {
        int ans =nums[0];
        for (int i = 1; i < nums.length; i++) {
            ans = ans ^ nums[i];
        }
        return ans;
    }
}

class Solution {
    public int singleNumber(int[] nums) {
        int ans = 0;
        for (int i = 0; i < 32; i++) {
            int count = 0;
            for (int num : nums) {
                count += (num >> i) & 1;
            }
            count %= 3;
            ans |= (count << i);
        }
        return ans;
    }
}

class Solution {
    public int rangeBitwiseAnd(int left, int right) {
        int cnt = 0;
        while(left != right){
            cnt++;
            left >>= 1;
            right >>= 1;
        }
        return left <<cnt;
    }
}