你要是智慧盈通的正在面试请别看!!
static List<int[]> allSorts = new ArrayList<int[]>()
public static void main(String[] args) {
int[] numArray = {1, 2, 3, 4, 5,8}
permutation(numArray, 0, numArray.length - 1)
int[][] a = new int[allSorts.size()][]
allSorts.toArray(a)
int count=0
// 打印验证
for (int i=0
int[] nums = a[i]
Boolean isShow = true
List<Integer> list = new ArrayList<>()
for (int j=0
if(j==3&&nums[j]==3){
isShow=false
break
}
if(j>1&&nums[j]==2&&nums[j-1]!=4){
isShow=false
break
}else if(j>1&&nums[j]==4&&nums[j-1]!=2){
isShow=false
break
}
list.add(nums[j])
}
if(isShow){
System.out.println(list)
count++
}
}
System.out.println("count"+count)
}
public static void permutation(int[] nums, int start, int end) {
if (start == end) { // 当只要求对数组中一个数字进行全排列时,只要就按该数组输出即可
int[] newNums = new int[nums.length]
for (int i=0
newNums[i] = nums[i]
}
allSorts.add(newNums)
} else {
for (int i=start
int temp = nums[start]
nums[start] = nums[i]
nums[i] = temp
permutation(nums, start + 1, end)
nums[i] = nums[start]
nums[start] = temp
}
}
}