Leetcode题解（20）

59. Spiral Matrix II

题目

这道题copy网上的代码

class Solution {
private:
    int step[4][2];
    bool canUse[100][100];
public:
    void dfs(int dep, vector<vector<int> > &matrix, int direct, int x, int y)
    {
        for(int i = 0; i < 4; i++)
        {
            int j = (direct + i) % 4;
            int tx = x + step[j][0];
            int ty = y + step[j][1];
            if (0 <= tx && tx < matrix.size() && 0 <= ty && ty < matrix[0].size() && canUse[tx][ty])
            {
                canUse[tx][ty] = false;
                matrix[tx][ty] = dep;              
                dfs(dep + 1, matrix, j, tx, ty);               
            }            
        }
    }
    
    vector<vector<int> > generateMatrix(int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        step[0][0] = 0;
        step[0][1] = 1;
        step[1][0] = 1;
        step[1][1] = 0;
        step[2][0] = 0;
        step[2][1] = -1;
        step[3][0] = -1;
        step[3][1] = 0;
        vector<vector<int> > ret(n, vector<int>(n));
        memset(canUse, true, sizeof(canUse));
        dfs(1, ret, 0, 0, -1);
        
        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 = 1;
        for(int i = 0 ; i < n; ++i) {
            nums[i] = i + 1;
            pCount *= (i + 1);
        }

        k--;
        string res = "";
        for(int i = 0 ; i < n; i++) {
            pCount = pCount/(n-i);
            int selected = k / pCount;
            res += ('0' + nums[selected]);
            
            for(int j = selected; j < n-i-1; j++)
                nums[j] = nums[j+1];
            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 = 0;
        while(temp != NULL)
        {
            count++;
            temp = temp->next;
        }
        k = k%count;
        if(k == 0)
            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;
    }
};