1. 题目

螺旋矩阵

2. 题意

给定一个二维数组，顺时针螺旋输出其中的元素。

3. 题解

3.1 自己想的

一层一层的，关键点在于如何进入下一循环。和确定停止的位置，找停止的位置写了个函数。

用控制比特位的方式来进行控制一次顺时针循环, 还被&优先级给坑到了。

class Solution {
public:vector<int> getEndPos(int x, int y) {if (x == 1)return {0, max(0,y - 1)};if (y == 1)return {max(0,x - 1), 0};if ( x == 2 || y == 2)return { 1, 0};int retx = 0, rety = 0;while ( x > 2 && y > 2) {retx++;rety++;x -= 2;y -= 2;}vector<int> tmp = getEndPos(x,y);retx += tmp[0];rety += tmp[1];return {retx, rety};}vector<int> spiralOrder(vector<vector<int>>& matrix) {vector<int> res;int cols = matrix[0].size();int rows = matrix.size();vector<int> cur{0,0};vector<int> ePos = getEndPos( rows, cols );std::cout << ePos[0] << ":" << ePos[1] << std::endl;int ctr_bit = 0;int lb = -1;int rb = cols;int ub = 0;int bb = rows;while (1) {res.push_back(matrix[cur[0]][cur[1]]);// std::cout << cur[1] << ":" << cur[0] << std::endl;if (cur == ePos) {break;}// printf("%d\n", ctr_bit);if ( ((ctr_bit & 1) == 0) && (cur[1] + 1 < rb ) ) {//printf("here");cur[1] = cur[1] + 1;continue;}ctr_bit |= 1;if ( (ctr_bit & 2) == 0 && (cur[0] + 1 < bb) ) {cur[0] = cur[0] + 1;continue;}ctr_bit |= 2;if ( (ctr_bit & 4) == 0 && ((cur[1] - 1) > lb) ){cur[1] = cur[1] - 1;printf("%d\n" ,cur[0]);continue;}ctr_bit |= 4;if ( (ctr_bit & 8) == 0 && cur[0] - 1 > ub) {cur[0] = cur[0] - 1;continue;}ctr_bit |= 8;++lb;--rb;++ub;--bb;ctr_bit &= 0;cur[1]++;}return res;}
};

3.2 大佬的代码

当成一圈一圈的，当上下边界或者左右边界重合的时候就停止。简洁易懂！

class Solution {
public:vector<int> spiralOrder(vector<vector<int>>& matrix) {vector<int> res;int lb = 0;int rb = matrix[0].size() - 1;int ub = 0;int db = matrix.size()  -1;while (1) {for ( int i = lb; i <= rb; ++i) {res.push_back(matrix[ub][i] );}if ( ++ub > db)break;for ( int i = ub; i <= db;++i) {res.push_back(matrix[i][rb]);}if (--rb < lb) {break;}for ( int i = rb; i >= lb; --i) {res.push_back(matrix[db][i]);}if ( --db < ub)break;for ( int i = db; i >= ub; --i) {res.push_back(matrix[i][lb]);}if ( ++lb > rb)break;}return res;}
};