腾讯精选50题—Day7题目54,59,61

腾讯精选50题—Day7题目54,59,61

  今天是刷题的第七天，元气满满的周一~~

1. 题目54 螺旋矩阵

(1) 题目描述

(2) 思路

  螺旋矩阵的问题比较简单，只要定好四个位置即可，lx、rx、by、ty，按照绿色箭头顺序遍历，在遍历的过程中保证lx不能大于rx同时by不能大于ty。

(3) 题解

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {

        int lx = 0;
        int rx = matrix[0].size() - 1;
        int by = 0;
        int ty = matrix.size() - 1;
        
        int number = matrix[0].size() * matrix.size();
        vector<int> result;

        while(true)
        {
            if (lx > rx || by > ty)
                break;
            for (int i = lx; i <= rx; i++)
                result.push_back(matrix[by][i]);
            by++;
            if (lx > rx || by > ty)
                break;
            for (int j = by; j <= ty; j++)
                result.push_back(matrix[j][rx]);
            rx--;
            if (lx > rx || by > ty)
                break;
            for (int k = rx; k >= lx; k--)
                result.push_back(matrix[ty][k]);
            ty--;
            if (lx > rx || by > ty)
                break;
            for (int t = ty; t >= by; t--)
                result.push_back(matrix[t][lx]);
            lx++;

        }
        return result;
    }
};

结果：

时间复杂度：$O(n^2)$
空间复杂度：$O(n^2)$

2. 题目59 螺旋矩阵II

(1) 题目描述

(2) 思路

  思路和54题类型，先声明一个空的矩阵，然后填数字。

(3) 题解

class Solution {
public:
    vector<vector<int>> generateMatrix(int n) {

        vector<vector<int>> result(n, vector<int>(n, 0));

        int lx = 0;
        int rx = n - 1;
        int by = 0;
        int ty = n - 1;

        int num = 1;
        while (true)
        {
            if (lx > rx || by > ty)
                break;
            for (int i = lx; i <= rx; i++)
            {
                result[by][i] = num;
                num++;
            }
            by++;
            if (lx > rx || by > ty)
                break;
            for (int j = by; j <= ty; j++)
            {
                result[j][rx] = num;
                num++;
            }
            rx--;
            if (lx > rx || by > ty)
                break;
            for (int k = rx; k >= lx; k--)
            {
                result[ty][k] = num;
                num++;
            }
            ty--;
            if (lx > rx || by > ty)
                break;
            for (int t = ty; t >= by; t--)
            {
                result[t][lx] = num;
                num++;
            }
            lx++;
        }

        return result;
    }
};

结果：

时间复杂度：$O(n^2)$
空间复杂度：$O(n^2)$

3. 题目61 旋转链表

(1) 题目描述

(2) 思路

  先成环，后断链。

(3) 题解

class Solution {
public:
    ListNode* rotateRight(ListNode* head, int k) {

        if (k == 0)
            return head;

        ListNode* p = head;
        ListNode* tail = NULL;
        int len = 0;
        while (p != NULL)
        {
            if (p->next == NULL)
                tail = p;

            p = p->next;
            len++;
            
        }
        if (len == 0) return NULL;
        int new_k = k % len;

        tail->next = head;
        p = tail;

        int move_step = len - new_k;

        while (move_step > 0)
        {
            p = p->next;
            move_step--;
        }
        ListNode* result = p->next;
        p->next = NULL;

        return result;
    }
};

结果：

时间复杂度：$O(n)$
空间复杂度：$O(1)$