Update Bits
题目：Given two 32-bit numbers, N and M, and two bit positions, i and j. Write a method to set all bits between i and j in N equal to M (e g , M becomes a substring of N located at i and starting at j)
Example
Given N=(10000000000)2, M=(10101)2, i=2, j=6
return N=(10001010100)2 Note
In the function, the numbers N and M will given in decimal, you should also return a decimal number. Challenge
Minimum number of operations? Clarification
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.
解题思路：
- 得到第i位到第j位的比特位为0，而其他位均为1的掩码mask。
- 使用mask与 N 进行按位与，清零 N 的第i位到第j位。
- 对 M 右移i位，将 M 放到 N 中指定的位置。
- 返回 N | M 按位或的结果。
Java代码
int updateBits(int n, int m, int i, int j) {
int ones = ~0;
int mask = 0;
if (j < 31) {
int left = ones << (j + 1);
int right = ((1 << i) - 1);
mask = left | right;
} else {
mask = (1 << i) - 1;
}
return (n & mask) | (m << i);
}
使用~0获得全1比特位，在j == 31时做特殊处理，即不必求left。求掩码的右侧1时使用了(1 << i) - 1, 题中有保证第i位到第j位足以容纳 M, 故不必做溢出处理。