day25

1.剑指 Offer 29. 顺时针打印矩阵

 方形盘龙问题，思路：在数组外面围一堵矩形的墙，碰到墙就掉头

代码：

#include <stdio.h>
#include <stdlib.h>
#define wall -1
#define a 0
int main() {
    int cs;
    int n,i,j,k=0,b,cnt,next;

    scanf("%d",&cs);
    while(cs--)
    {

        scanf("%d",&n);
        int s[32][32];
        for(i=0; i<n+1; i++) {
            s[0][i]=wall;
            s[i][0]=wall;
            s[n+1][i]=wall;
            s[i][n+1]=wall;
        }
        for(i=1; i<=n; i++) {
            for(j=1; j<=n; j++) {
                s[i][j]=a;
            }
        }
        int di[]= {0,1,0,-1};
        int dj[]= {1,0,-1,0};
        b=n*n;
        i=1;
        j=0;
        k=0;
        cnt=0;
        while(cnt<b) {
            next=s[i+di[k]][j+dj[k]];
            if(next==wall||next!=a)
                k=(k+1)%4;
            i=i+di[k];
            j=j+dj[k];
            cnt++;
            s[i][j]=cnt;
        }
        for(i=1; i<=n; i++) {
            for(j=1; j<=n; j++) {
                printf("%4d",s[i][j]);
            }
            printf("\n");
        }

    }
    return 0;
}

 1）顺时针打印矩阵代码（方形盘龙思路）：

class Solution {
public:
    #define wall -1
    int next[4][2] ={0,1,1,0,0,-1,-1,0};//i = i + next[][0],j = j + next[][1]
    vector<int> res;
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
       int m = matrix.size();
       if(m == 0) return res;
       int n = matrix[0].size();
       if(n == 0) return res;
       if(m == 1){
           for(int col = 0;col < n;col ++)
            res.push_back(matrix[0][col]);
           return res;
       }
       if(n == 1){
           for(int row = 0;row < m;row ++)
            res.push_back(matrix[row][0]);
           return res;
       }
       FirstWall(matrix,m,n);
       int i = 1,j = 1,k = 0;
       int cnt = m * n - (2 * (m + n - 2));
       while(cnt --){
           if(matrix[i + next[k][0]][j + next[k][1]] == wall){
               k = (k + 1) % 4;
           }
           res.push_back(matrix[i][j]);
           matrix[i][j] = wall;
           i = i + next[k][0];
           j = j + next[k][1];
       }
       return res;
    }

private:
    void FirstWall(vector<vector<int>>& matrix,int m,int n){ //第一层也就是最外面一层“墙”
        for(int j = 0;j <= n - 2;j ++){
           res.push_back(matrix[0][j]);printf("%d ",matrix[0][j]);
           matrix[0][j] = wall;
       }
       for(int i = 0;i <= m - 2;i ++){
           res.push_back(matrix[i][n - 1]);printf("%d ",matrix[i][n - 1]);
           matrix[i][n - 1] = wall;
       }
       for(int j = n - 1;j >= 1;j --){
           res.push_back(matrix[m - 1][j]);printf("%d ",matrix[m - 1][j]);
           matrix[m - 1][j] = wall;
       }
       for(int i = m - 1;i >= 1;i --){
           res.push_back(matrix[i][0]);printf("%d ",matrix[i][0]);
           matrix[i][0] = wall;
       }
    }
};

 2）https://leetcode.cn/leetbook/read/illustration-of-algorithm/5vfb5v/

t：top，r：right，b：below，l：left

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
       vector<int> res;
       if(matrix.empty())  return res;
       int t = 0,r = matrix[0].size() - 1,b = matrix.size() - 1,l = 0;
       while(1){
        for(int i = l;i <= r;i ++)  res.push_back(matrix[t][i]); //from left to right(the top row)
        if(++ t > b) break;
        for(int i = t;i <= b;i ++)  res.push_back(matrix[i][r]); //from top to below(the right col)
        if(-- r < l) break;
        for(int i = r;i >= l;i --)  res.push_back(matrix[b][i]); //from right to left(the below row)
        if(-- b < t) break;
        for(int i = b;i >= t;i --)  res.push_back(matrix[i][l]); //from below to top(the left col) 
        if(++ l > r) break;
       }
       return res;
    }
};

2.剑指 Offer 31. 栈的压入、弹出序列

 https://leetcode.cn/leetbook/read/illustration-of-algorithm/5wxgft/

class Solution {
public:
    bool validateStackSequences(vector<int>& pushed, vector<int>& popped) {
       stack<int> stk;
       int i = 0;
       for(auto x : pushed){
           stk.push(x);
           while(! stk.empty() && stk.top() == popped[i]){
             stk.pop();
             i ++;
           }
       }
       return stk.empty();
    }
};