Lintcode: Update Bits
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) Have you met this question in a real interview? Yes 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.
以题中例子为例,做一个滤波器在i,j之间:11110000011,来跟N按位与,再把M左移i位,按位或
1 class Solution { 2 /** 3 *@param n, m: Two integer 4 *@param i, j: Two bit positions 5 *return: An integer 6 */ 7 public int updateBits(int n, int m, int i, int j) { 8 // write your code here 9 int len = j-i+1; 10 int temp = 0; 11 for (int x=0; x<len; x++) { 12 temp |= 1<<x; 13 } 14 temp = ~(temp<<i); 15 return (n&temp) | (m<<i); 16 } 17 }