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Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.
Example : Given a = 1 and b = 2, return 3.
class Solution { public: int getSum(int a, int b) { return b == 0 ? a : getSum(a ^ b, (a & b) << 1); } };https://discuss.leetcode.com/topic/50315/a-summary-how-to-use-bit-manipulation-to-solve-problems-easily-and-efficiently 补充位运算符相关知识： 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 Examples
//Count the number of ones in the binary representation of the given number int count_one(int n) { while(n) { n = n&(n-1); count++; } return count; } //Is power of four (actually map-checking, iterative and recursive methods can do the same) bool isPowerOfFour(int n) { return !(n&(n-1)) && (n&0x55555555); //check the 1-bit location; }