Java 求最小+;一个大于给定数字且与给定整数具有相同二进制权重的ve整数
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,找到与N具有相同二进制权重的大于N的最小整数Java 求最小+;一个大于给定数字且与给定整数具有相同二进制权重的ve整数,java,algorithm,bit-manipulation,time-complexity,permutation,Java,Algorithm,Bit Manipulation,Time Complexity,Permutation,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,找到与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 ;
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;
}
公共静态整数计算(整数){
int count=0,下一个编号;
char[]arr=Integer.tobinarysting(number.toCharArray();
对于(int i=0;i
我的解决方案很好,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?您可以使用算法进行下一次词典排列。中整数数组的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;
}
布尔nextPermutation(int[]数组){
//查找最长的非递增后缀
int i=array.length-1;
而(i>0&&array[i-1]>=array[i])
我--;
//现在i是后缀的头索引
//我们已经是最后一个排列了吗?
if(i找到了一种方法来实现相同的O(logn)复杂度
主要步骤:-
将数字转换为二进制字符数组。(非常简单)
根据二进制数组的模式对其进行分类。(在注释中解释)
根据类别调整二进制数组的0和1
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;
}
公共静态整数计算(整数){
int位模式类别、结果;
char[]charArr=Integer.toBinaryString(number.toCharArray();
bitPatternCategory=determineNumberType(charArr);
结果=findNextDesired(bitPatternCategory,charArr);
返回结果;
}
公共静态int findnextrequired(int bitPatternCategory,char[]arr){
整数;
//系统输出打印(“bitPatternCategory\t”+bitPatternCategory+“\n”);
字符[]临时=新字符[arr.length+1];
如果(bitPatternCategory==0){
temp=getNextForCat0(arr,temp);
}else if(bitPatternCategory==1){
temp=getNextForCat1(arr,temp);
}否则{
温度=getNextForCat2(arr);
}
number=Integer.parseInt(String.valueOf(temp),2);
返回号码;
}
私有静态字符[]getNextForCat2(字符[]arr){
//对于除0类和1类以外的所有情况,例如11000、1001、1101010
//从RHS开始查找1后第一次出现的0,交换这些位,然后移动
//阵列的所有剩余1(RHS到交换位)到RHS
int j=0,反坡向=0;
布尔标志=假;
对于(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;
打破
}
}
而((j0){
arr[arr.length-counterforone]=“1”;
反褶孔--;
}
返回arr;
}
私有静态字符[]getNextForCat1(字符[]arr,字符[]temp){
//对于eg 11100的所有位都在LHS上的情况,则添加一个值为1和的新位
//将剩余1的起点从第2个1移到RHS,使1111变为10111
int j=1,反坡向=0;
而((j0){
温度[temp.length Counterforone]=“1”;
反褶孔--;
}
arr=温度;
返回arr;
}
私有静态char[]getNextForCat0(char[]arr,char[]temp){
//对于所有位仅为1的情况,则添加一个值为1和的新位
//将剩余1的起点从第2个1移到RHS,使1111变为10111
对于(int i=0;i=1;--i)
温度[i]=温度[i-1];
温度[1]=“0”;
arr=温度;
返回arr;
}
公共静态int determinuteNumberType(char[]arr){
int stateMachine=0;//类别0适用于例如111、1111的所有类别
//1类为1和0 11100、110000
//第2类为混合型,或者我们可以说剩余的组合10111101
//假设MSB始终为1
字符温度=arr[0];
对于(int i=0;i
此操作有时称为“snoob”。下面是一个.Prob
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
}