【C++刷题笔记】螺旋矩阵的两种写法

螺旋矩阵有两种判断大循环结束的方式，第一种是判断需要循环多少次，奇数的话需要额外处理；第二种通过取多少个数判断，不需要额外处理

方法一：

class Solution {public:vector<int> spiralOrder(vector<vector<int>>& matrix) {vector<vector<int>> &v = matrix;vector<int> ret;int x = 0, y = 0; // x表示当前这一圈的最左上角的纵向起始坐标，y表示横向起始坐标int w = v[0].size();// w表示矩阵当前最左边的边界int h = v.size(); // h表示矩阵做下面的边界int n = min(w,h) / 2;int h1 = h;int w1 = w;while (n--) {int i = x, j = y; // i表示纵向坐标，j表示横向坐标for (; j < w; ++j) {// cout << v[i][j] << " ";ret.push_back(v[i][j]);}--j;++i;for ( ; i < h; ++i) {// cout << v[i][j] << " ";ret.push_back(v[i][j]);}--i;--j;for (; j >= y; --j) {// cout << v[i][j] << " ";ret.push_back(v[i][j]);}++j;--i;for ( ; i > x; --i) {// cout << v[i][j] << " ";ret.push_back(v[i][j]);}x = x + 1;y = y + 1;w -= 1;h -= 1;}if (w1 < h1 && w1 %2 == 1) {for (int i = x; i < h; ++i) {// cout << v[i][y] << " ";ret.push_back(v[i][y]);}}if (h1 < w1 && h1 % 2 == 1) {for (int i = y; i < w; ++i) {// cout << v[x][i] << " ";ret.push_back(v[x][i]);}}if (h1 == w1 && h1 %2 == 1) {// cout << v[x][y] << " ";ret.push_back(v[x][y]);}return ret;// cout << endl;}};

第二种：

#include "iostream"using namespace std;class Solution {public:vector<int> spiralOrder(vector<vector<int>>& matrix) {vector<int> ret;int m = matrix.size();int n = matrix[0].size();int x = 0, y = 0;int startX = 0, startY = 0;int count = 1;int sum = matrix.size() * matrix[0].size();while (count <= sum) {x = startX, y = startY;while (count <= sum && y < n) {ret.push_back(matrix[x][y]);count++;y++;}y--;x++;while (count <= sum && x < m) {ret.push_back(matrix[x][y]);x++;count++;}x--;y--;while (count <= sum && y >= startY) {ret.push_back(matrix[x][y]);y--;count++;}y++;x--;while (count <= sum && x > startX) {ret.push_back(matrix[x][y]);x--;count++;}m--;n--;startX = startX + 1;startY = startY + 1;}return ret;}};