Permutation Sequence

The set [1,2,3,…,n] contains a total of n! unique permutations.

By listing and labeling all of the permutations in order,
We get the following sequence (ie, for n = 3):

1. "123"
2. "132"
3. "213"
4. "231"
5. "312"
6. "321"

Given n and k, return the kth permutation sequence.

Note: Given n will be between 1 and 9 inclusive.

public static String getPermutation(int n, int k) {
//数组赋值
int num[]=new int[n];
   for(int i=0;i     num[i]=i+1;
   if(n==1){
    if(k==1)
    return "1";
    else {
return "";
}
   }
   int left=n-2;
   //遍历k次
   for(int count=1;count     left=n-2;
    //找left 从最后一个起第一个这个值比后一个值小
    while(left>=0&&num[left]>num[left+1])
    left--;
    //不能遍历了
    if(left<0)
    break;
    int mincount=1;
    int right=left;
    int min=num[left];
    //找right值 left后面比left值大的最小值
    for(int i=left;i     if(num[i]>num[left]){
    if(mincount==1){
    right=i;
    min=num[i];
    mincount=2;
    }
    else {
if(num[i] right=i;
    min=num[i];
}

}
   
    }
   
   
    }
    //交换left和right
    int temp=num[left];
    num[left]=num[right];
    num[right]=temp;
    int j=n-1;
    //left后面值倒序
    for(int i=left+1;i     int temp1=num[i];
    num[i]=num[j];
    num[j]=temp1;
    }
   }
   if(left<0)
    return "";
   //当+多的时候使用StringBuffer性能高
   StringBuffer sBuffer=new StringBuffer("");
   for(int i=0;i     sBuffer.append(num[i]);
   return sBuffer.toString();
   
}