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190. Reverse Bits

时间:2015-04-08 07:55:03      阅读:150      评论:0      收藏:0      [点我收藏+]

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Reverse bits of a given 32 bits unsigned integer.

For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as00111001011110000010100101000000).

Follow up:
If this function is called many times, how would you optimize it?

Related problem: Reverse Integer

两个指针i,j从前后向中间扫,如果bit[i] != bit[j],反转bit[i]和bit[j].

public class Solution {
  // you need treat n as an unsigned value
  public int reverseBits(int n) {
    for (int i = 0; i < 16; i++) {
      n = swapBit(n, i, 31-i);
    }
    return n;
  }

  public int swapBit(int n, int i, int j) {
    int low = (n >> i) & 1;
    int high = (n >> j) & 1;
    if ((low ^ high) == 1) {
      n ^= 1 << i;
      n ^= 1 << j;
    }
    return n;
  }
}

190. Reverse Bits

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原文地址:http://www.cnblogs.com/shini/p/4401345.html

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