我经历了两种和三种颜色的解决方案，但我无法获得四种颜色。
请帮忙。
是rrbb????yyyggg吗？
我们如何交换绿旗？
我尝试了下面的解决方案，但它无法使用绿色交换最后一个黄色。
public class count123 {
// Java program to sort an array of 0, 1 and 2,3
static void sort0123(int a[], int arr_size)
{
int lo = 0;
int hi = arr_size - 1;
int mid = 0,temp=0;
int h2=arr_size - 1;
while (mid <= hi)
{
switch (a[mid])
{
case 0:
{
temp = a[lo];
a[lo] = a[mid];
a[mid] = temp;
lo++;
mid++;
break;
}
case 1:
mid++;
break;
case 2:
{
temp = a[mid];
a[mid] = a[hi];
a[hi] = temp;
hi--;
h2=hi;
break;
}
case 3:{
temp = a[mid];
a[mid] = a[h2];
a[h2] = temp;
// h2--;
//hi=h2;
break;
}
}
}
}
/* Utility function to print array arr[] */
static void printArray(int arr[], int arr_size)
{
int i;
for (i = 0; i < arr_size; i++)
System.out.print(arr[i]+" ");
System.out.println("");
}
/*Driver function to check for above functions*/
public static void main (String[] args)
{
int arr[] = {0, 1, 0,1,2,2,0,3,3,0,0,1};
int arr_size = arr.length;
sort0123(arr, arr_size);
System.out.println("Array after seggregation ");
printArray(arr, arr_size);
}
}
/*This code is contributed by Devesh Agrawal*/
答案 0 :(得分：2)
我运行你的代码并意识到你的代码进入了无限循环，这让你的程序什么都不做。
在主要方法中，sort0123(arr, arr_size);被调用，在此方法中，while (mid <= hi)，mid = 6和hi = 9，这意味着6 <= 9，由于此原因，此条件返回无限真实，因为它总是在某个时刻到达案例3而在案例3中，mid和hi的值不会改变。
为了能够运行您的代码，您应该逻辑地更改mid和/或hi的值，并且使用它，您的程序可以按照您想要的方式运行。
答案 1 :(得分：1)
关于荷兰国旗聚合的观点是始终保持不变量。处于诸如0000 ... 11..XXX..222
的状态对于4色变化，假设您按照代码0,1,3,2的顺序进行排序，则需要对规则进行如下调整：
遵循上述规则，需要通过以下方式更改代码：
while (mid <= h2)
。 。
case 2:
{
temp = a[mid];
a[mid] = a[hi];
a[hi] = temp;
hi--;
h2--;
break;
}
case 3:{
temp = a[mid];
a[mid] = a[h2];
a[h2] = temp;
h2--;
break;
}
按顺序排序0,1,2,3只需交换'2'和'3'的case语句。
PS：请将问题描述为帮助人们轻松理解问题。
答案 2 :(得分：1)
#include <iostream>
using namespace std;
void DutchNationalFlag4(int * arr , int n )
{
int low = 0 , mid1 = 0 , mid2 = n- 1 , high = n-1 ;
int temp;
// This type of problems are best solved by using invariants
// arr[ 0 .. low -1 ] conatins 0
// arr[ low .. mid1 -1 ] contains 1
// arr[ mid1 .. mid2 ] is unknown
// arr[ mid2 + 1 ... high - 1 ] contains 2
// arr[ high .. n-1 ] contains 3
// termination condition : Evert Iteration unkown length decreases so eventually will be Zero
while( mid1 <= mid2 )
{
switch( arr[ mid1 ])
{
case 0 :
{
std::swap( arr[ low ] , arr [ mid1 ] );
low ++ ;
mid1 ++;
break;
}
case 1 :
{
mid1 ++;
break;
}
case 2 :
{
std::swap( arr[ mid1 ] , arr[ mid2 ]);
mid2 -- ;
break;
}
case 3 :
{
std::swap( arr[ mid1 ] , arr[ mid2 ]);
std::swap( arr[ mid2 ] , arr[ high ]);
mid2 --;
high --;
break;
}
}
}
for( int i =0 ; i < n ; i++)
{
cout << arr[ i ] << " " ;
}
}
int main()
{
int arr[] = {1,2,3,0,2,1,3,0,2,1,0,1,3,1,0,2,1,0};
int n = sizeof(arr) / sizeof(arr[0]) ;
DutchNationalFlag4(arr , n );
return 0;
}
我已经在上面的链接中解决了！ 干杯
在问问题时，请尽量保持清楚。还要确保您的变量是直观命名的，或者您提供了适当的注释。 坦白说，我对hi和h2感到困惑，中点！
答案 3 :(得分：0)
这是我用python编写的解决方案。其背后的原理是将左右中间子数组之间的第四种颜色（中间右颜色）压缩。它随着算法的进行定义颜色。
它的时间复杂度为O（n），空间复杂度为O（1）
from typing import List
def dutch_variant_four_colors(array: List[int]) -> List[int]:
left = array[0]
mid_left = None
right = None
left_i = 0
mid_left_i = 0
mid_right_i = len(array)
right_i = len(array)
while mid_left_i < right_i and mid_left_i < mid_right_i:
if (array[mid_left_i] == left):
array[mid_left_i], array[left_i] = array[left_i], array[mid_left_i]
mid_left_i += 1
left_i += 1
elif (right is None or array[mid_left_i] == right):
right_i -= 1
mid_right_i = right_i
array[mid_left_i], array[right_i] = array[right_i], array[mid_left_i]
right = array[right_i]
else: # it is a mid value
if (mid_left is None):
mid_left = array[mid_left_i]
if (array[mid_left_i] == mid_left):
mid_left_i += 1
else:
mid_right_i -= 1
array[mid_left_i], array[mid_right_i] = array[mid_right_i], array[mid_left_i]
return array
# Sample usages
print(dutch_variant_four_colors([1,1,3,3,4,3,2,2]))
print(dutch_variant_four_colors([1,2,3,4,2,3,1,3]))
print(dutch_variant_four_colors([1,2,3,4,4]))
print(dutch_variant_four_colors([0,1,2,5,5,2,2,0]))
print(dutch_variant_four_colors([1,0,3,0,5,5]))
print(dutch_variant_four_colors([1,2,3,2,5,1,1,3]))
print(dutch_variant_four_colors([5,1,2,3,2,1,1,3]))
print(dutch_variant_four_colors([3,2,5,1]))
print(dutch_variant_four_colors([3,2,1,5]))
print(dutch_variant_four_colors([3,2,1,3,5,2,2,1,2,3,5,3,2,1,3,5,3,3,2,2,2,5,5,5,3,3,2,5,3,1,2,3,2,1,3,2,1,1,2,3,2,3,2,1,2,3,2,1,2,3,2,2,2,2,2,3,3,3,1,2,2,1,1,2,3]))
要约在https://gist.github.com/lopespm/81a336871ce6074f63f3cad349c3a95d上可用
答案 4 :(得分：0)
//Dutch National Flag for N different values Optimized
//O(n) optimized solution.
//O(1) space needed where x is distinct colors (it will not change as no. of values increases)
public static void DutchNationalFlagforNvaluesOptimized(int distinctFlags, int[] flags)
{
int high = distinctFlags - 1, mid = distinctFlags - 2;
int[] indexArr = new int[distinctFlags];
for (int i = 0; i < high; i++)
{
indexArr[i] = 0;
}
indexArr[high] = flags.Length - 1;// index Array is distinct flags indexes
for (; indexArr[mid] <= indexArr[high]; )
{
if (flags[indexArr[mid]] == high)
{
int temp = flags[indexArr[high]];
flags[indexArr[high]] = flags[indexArr[mid]];
flags[indexArr[mid]] = temp;
indexArr[high]--;
}
else
if (flags[indexArr[mid]] == mid)
{
indexArr[mid]++;
}
else
{
int currentMidValue = flags[indexArr[mid]];
for (int i = mid; i > currentMidValue; i--)
{
int temp = flags[indexArr[i]];
flags[indexArr[i]] = flags[indexArr[i - 1]];
flags[indexArr[i - 1]] = temp;
indexArr[i]++;
}
indexArr[currentMidValue]++;
}
}
for (int i = 0; i < flags.Length; i++)
{
System.Console.Write(flags[i] + ", ");
}
}
答案 5 :(得分：0)
let a:string[] = ['1','2','1','0','2','4','3','0','1','3'];
function sort3(a:string[]):void{
let low = 0;
let mid1 = 0;
let mid2 = 0;
let mid3 = 0;
let high = a.length - 1;
while(mid3<=high){
switch(a[mid3]){
case '0': [a[mid3],a[low]] = [a[low],a[mid3]];
low++;
if(mid1 < low)
mid1++;
if(mid2 < mid1)
mid2++;
if(mid3 < mid2)
mid3++;
break;
case '1': [a[mid3],a[mid1]] = [a[mid1],a[mid3]];
mid1++;
if(mid2 < mid1)
mid2++;
if(mid3 < mid2)
mid3++
break;
case '2': [a[mid2],a[mid3]] = [a[mid3],a[mid2]];
mid2++;
mid3++;
break;
case '3':
mid3++;break;
case '4': [a[mid3],a[high]] = [a[high],a[mid3]];
high--;
}
}
}