Leetcode题解（20）

59. Spiral Matrix II

题目

这道题copy网上的代码

 class Solution {
 private:
     int step[][];
     bool canUse[][];
 public:
     void dfs(int dep, vector<vector<int> > &matrix, int direct, int x, int y)
     {
         for(int i = ; i < ; i++)
         {
             int j = (direct + i) % ;
             int tx = x + step[j][];
             int ty = y + step[j][];
             if ( <= tx && tx < matrix.size() &&  <= ty && ty < matrix[].size() && canUse[tx][ty])
             {
                 canUse[tx][ty] = false;
                 matrix[tx][ty] = dep;
                 dfs(dep + , matrix, j, tx, ty);
             }
         }
     }

     vector<vector<int> > generateMatrix(int n) {
         // Start typing your C/C++ solution below
         // DO NOT write int main() function
         step[][] = ;
         step[][] = ;
         step[][] = ;
         step[][] = ;
         step[][] = ;
         step[][] = -;
         step[][] = -;
         step[][] = ;
         vector<vector<int> > ret(n, vector<int>(n));
         memset(canUse, true, sizeof(canUse));
         dfs(, ret, , , -);

         return ret;
     }

 };

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60. Permutation Sequence

题目

分析：这道题主要考数学推断

在n!个排列中，第一位的元素总是(n-1)!一组出现的，也就说如果p = k / (n-1)!，那么排列的最开始一个元素一定是nums[p]。

假设有n个元素，第K个permutation是
a1, a2, a3, .....   ..., an
那么a1是哪一个数字呢？
那么这里，我们把a1去掉，那么剩下的permutation为
a2, a3, .... .... an, 共计n-1个元素。 n-1个元素共有(n-1)!组排列，那么这里就可以知道
设变量K1 = K
a1 = K1 / (n-1)!
同理，a2的值可以推导为
a2 = K2 / (n-2)!
K2 = K1 % (n-1)!
 .......
a(n-1) = K(n-1) / 1!
K(n-1) = K(n-2) /2!
an = K(n-1)

代码如下：

 class Solution {
 public:
     string getPermutation(int n, int k) {
         vector<int> nums(n);
         int pCount = ;
         for(int i =  ; i < n; ++i) {
             nums[i] = i + ;
             pCount *= (i + );
         }

         k--;
         string res = "";
         for(int i =  ; i < n; i++) {
             pCount = pCount/(n-i);
             int selected = k / pCount;
             res += ('' + nums[selected]);

             for(int j = selected; j < n-i-; j++)
                 nums[j] = nums[j+];
             k = k % pCount;
         }
         return res;
     }
 };

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61. Rotate List

题目

代码如下：

 /**
  * Definition for singly-linked list.
  * struct ListNode {
  *     int val;
  *     ListNode *next;
  *     ListNode(int x) : val(x), next(NULL) {}
  * };
  */
 class Solution {
 public:
     ListNode* rotateRight(ListNode* head, int k) {
         //如果k值大于链表长度应该怎么处理
         if(NULL == head)
             return NULL;
         ListNode* temp = head;
         int count = ;
         while(temp != NULL)
         {
             count++;
             temp = temp->next;
         }
         k = k%count;
         if(k == )
             return head;
         ListNode *first,*second;
         first = head;
         int i=k;
         while(i--)
         {
             //if(NULL == first)
             //    return NULL;
             first = first->next;
         }
         second = head;
         while(first->next != NULL)
         {
             first = first->next;
             second = second->next;
         }
         temp = head;
         head = second->next;
         first->next = temp;
         second->next = NULL;
         return head;
     }
 };