lintcode-179-更新二进制位

179-更新二进制位

给出两个32位的整数N和M，以及两个二进制位的位置i和j。写一个方法来使得N中的第i到j位等于M（M会是N中从第i为开始到第j位的子串）

注意事项

In the function, the numbers N and M will given in decimal, you should also return a decimal number.

说明

You can assume that the bits j through i have enough space to fit all of M. That is, if M=10011， you can assume that there are at least 5 bits between j and i. You would not, for example, have j=3 and i=2, because M could not fully fit between bit 3 and bit 2.

样例

给出N = (10000000000)2，M = (10101)2， i = 2, j = 6
返回 N = (10001010100)2

挑战

最少的操作次数是多少？

标签

比特位操作 Cracking The Coding Interview

思路

先将 N 的 i 到 j 位清零，把 M 向右移动 i 位，然后将 N 与上 M

code

class Solution {
public:
    /**
     *@param n, m: Two integer
     *@param i, j: Two bit positions
     *return: An integer
     */
    int updateBits(int n, int m, int i, int j) {
        // write your code here
        for (int pos = i; pos <= j; ++pos) {
            n &= ~(1 << pos);
        }
        m <<= i;
        return n | m;
    }
};