LeetCode Bit Manipulation 技巧总结

https://leetcode.com/problems/sum-of-two-integers/discuss/84278/A-summary%3A-how-to-use-bit-manipulation-to-solve-problems-easily-and-efficiently

Six common bit operator

• &
• |
• ~
• ^
• >>
• <<

Set union A | B
Set intersection A & B
Set subtraction A & ~B
Set negation ALL_BITS ^ A or ~A
Set bit A |= 1 << bit
Clear bit A &= ~(1 << bit)
Test bit (A & 1 << bit) != 0
Extract last bit A&-A or A&~(A-1) or x^(x&(x-1))
Remove last bit A&(A-1)
Get all 1-bits ~0

some examples:
count the number of 1s of a given number

int count_one(int n) {
    while(n) {
        n = n&(n-1);
        count++;
    }
    return count;
}

check the power of 4

bool isPowerOfFour(int n) {
    return !(n&(n-1)) && (n&0x55555555);
    //check the 1-bit location;
}

sum of two integers:

int getSum(int a, int b) {
    return b==0? a:getSum(a^b, (a&b)<<1); //be careful about the terminating condition;
}

return the missing number

int missingNumber(vector<int>& nums) {
    int ret = 0;
    for(int i = 0; i < nums.size(); ++i) {
        ret ^= i;
        ret ^= nums[i];
    }
    return ret^=nums.size();
}

Find the largest power of 2 (most significant bit in binary form), which is less than or equal to the given number N.

long largest_power(long N) {
    //changing all right side bits to 1.
    N = N | (N>>1);
    N = N | (N>>2);
    N = N | (N>>4);
    N = N | (N>>8);
    N = N | (N>>16);
    return (N+1)>>1;
}

…and many more other problems