求大于给定数且与给定整数具有相同二进制权重的最小ve整数的算法

Question

正整数的二进制权重是其二进制表示中1的个数。例如，十进制数1的二进制权重为1，十进制数7（二进制为111）的二进制权重为3。

给定一个正整数N，找出与N具有相同二进制权重的大于N的最小整数。

public static int compute(int number)   {
    int  count = 0, nextNumber;
    char[] arr =  Integer.toBinaryString(number).toCharArray();
    for(int i =0 ; i < arr.length ;++i) {
        if(arr[i] == '1')
            ++count;
    }
    nextNumber = findNextNumber(number,count);
    return nextNumber;
}
public static int findNextNumber(int number, int weight) {
    char[] arr;
    boolean flag = true;
    int count;
    while (flag) {
        // increment number and convert it into char array
        arr = Integer.toBinaryString(++number).toCharArray();
        count = 0;
        for(int i =0 ; i < arr.length; ++i) {
            if(arr[i] == '1')
                ++count;
        }
        if(count == weight) {
            flag = false;
        }
    }

    return number;
}

我的解决方案运行良好，但它的复杂性似乎是O（NlogN）。这可以通过其他方法在O（N）或O（logn）中实现吗？

Answer 1

找到了一种在O（logn）复杂度上实现相同的方法。

主要步骤：-

1. 将数字转换为二进制字符数组。（非常简单）
2. 根据模式对二进制数组进行分类。（在评论中解释）
3. 根据类别调整二进制数组的0和1
4. 将二进制字符数组转换回数字

代码：-

public static int compute(int number)   {
    int bitPatternCategory , result;
    char[] charArr = Integer.toBinaryString(number).toCharArray();
    bitPatternCategory = determineNumberType(charArr);
    result = findNextDesired(bitPatternCategory, charArr);
    return result;
}
public static int findNextDesired(int bitPatternCategory, char[] arr) {
    int number;
    //System.out.print("bitPatternCategory\t"+bitPatternCategory+"\n");
    char[] temp = new char[arr.length + 1];
    if (bitPatternCategory == 0) {
        temp = getNextForCat0(arr, temp);
    } else if (bitPatternCategory == 1) {
        temp = getNextForCat1(arr, temp);
    } else {
        temp = getNextForCat2(arr);
    }
    number = Integer.parseInt(String.valueOf(temp),2);
    return number;
}

private static char[] getNextForCat2(char[] arr) {
    // for all cases except cat 0 and 1, for example 110011000,  1001, 1101010
    //  Find first occurrence of 0 after 1 starting from RHS, exchange these bits and then move
    // all remaining 1's(on RHS to the exchanged bits) to RHS of the array
    int j =0,counterForOnes = 0;
    boolean flag = false;
    for (int i = arr.length - 1; i >= 0; --i) {
        if ((arr[i] == '1') && (flag == false)) {
            flag = true;
        } else if ((arr[i] == '0') && (flag == true)) {
            char tempChar = arr[i];
            arr[i] = arr[i + 1];
            arr[i + 1] = tempChar;
            j = i+2;
            break;
        }
    }
    while((j < arr.length) && (arr[j]!='0'))    {
        arr[j] = '0';
        ++counterForOnes;
        ++j;
    }
    while(counterForOnes > 0) {
        arr[arr.length-counterForOnes]= '1';
        counterForOnes--;
    }
    return arr;
}

private static char[] getNextForCat1(char[] arr, char[] temp) {
    // for cases when all ones are on LHS for eg 11100, then add a new bit with value 1 and
    //  shift remaining 1's start from 2nd 1 towards RHS, so 1111 become 10111
    int j = 1,counterForOnes= 0;
    while((j < arr.length) && (arr[j]!='0'))    {
        arr[j] = '0';
        ++counterForOnes;
        ++j;
    }
    for (int i = 0; i < arr.length; ++i) {
        temp[i] = arr[i];
    }
    temp[temp.length-1] = '0';
    while(counterForOnes > 0) {
        temp[temp.length-counterForOnes]= '1';
        counterForOnes--;
    }
    arr = temp;
    return arr;
}

private static char[] getNextForCat0(char[] arr, char[] temp) {
    // for cases when all bits are ones only, then add a new bit with value 1 and
    //  shift remaining 1's start from 2nd 1 towards RHS, so 1111 become 10111
    for (int i = 0; i < arr.length; ++i) {
        temp[i] = arr[i];
    }
    for (int i = temp.length - 1; i >= 1; --i)
        temp[i] = temp[i - 1];
    temp[1] = '0';
    arr = temp;
    return arr;
}

public static int determineNumberType(char[] arr)   {
    int stateMachine = 0;   //Category 0 for all ones for eg 111, 1111
                            //Category 1 for ones and zeroes  11100, 110000
                            //Category 2 for mix ones or we can say remaining combinations 1011, 11101
    // Assuming MSB will always be 1
    char temp = arr[0];
    for(int i = 0 ; i < arr.length ; ++i)   {
        if((temp == arr[i]) && (stateMachine == 0)) {
            stateMachine = 0;
        } else if(((temp != arr[i])) && (stateMachine == 0))    {
            stateMachine = 1;
            temp = arr[i];
        }else if(((temp == arr[i])) && (stateMachine == 1)) {
            stateMachine = 1;
        }else if(((temp != arr[i])) && (stateMachine == 1)) {
            stateMachine = 2;
            //temp = arr[i];
            break;
        }
    }   
    return stateMachine;
}

Answer 2

可以对下一个字典排列使用算法。整型数组的Java实现。只需调整它以使用位而不是数组项：

boolean nextPermutation(int[] array) {
    // Find longest non-increasing suffix
    int i = array.length - 1;
    while (i > 0 && array[i - 1] >= array[i])
        i--;
    // Now i is the head index of the suffix

    // Are we at the last permutation already?
    if (i <= 0)
        return false;

    // Let array[i - 1] be the pivot
    // Find rightmost element that exceeds the pivot
    int j = array.length - 1;
    while (array[j] <= array[i - 1])
        j--;
    // Now the value array[j] will become the new pivot
    // Assertion: j >= i

    // Swap the pivot with j
    int temp = array[i - 1];
    array[i - 1] = array[j];
    array[j] = temp;

    // Reverse the suffix
    j = array.length - 1;
    while (i < j) {
        temp = array[i];
        array[i] = array[j];
        array[j] = temp;
        i++;
        j--;
    }

    // Successfully computed the next permutation
    return true;
}

Answer 3

这种操作有时被称为“snoob”。以下是《黑客的快乐书》中的一系列方法。也许最好使用整数。可能编译为硬件指令（未测试）的跟踪零数：

int snoob1(int x) {
   int smallest, ripple, ones;  // x = xxx0 1111 0000
   smallest = x & -x;           //     0000 0001 0000
   ripple = x + smallest;       //     xxx1 0000 0000
   ones = x ^ ripple;           //     0001 1111 0000
   ones = ones >>> (2 + Integer.numberOfTrailingZeros(x)); //     0000 0000 0111
   return ripple | ones;        //     xxx1 0000 0111
}

（由于Java中的移位计数是模32，所以“2”部分可能存在溢出问题）