给定一个包含?m x n?个元素的矩阵(m 行, n 列),请按照顺时针螺旋顺序,返回矩阵中的所有元素。
示例?1:
输入:
[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]
输出: [1,2,3,6,9,8,7,4,5]
示例?2:
输入:
[
?[1, 2, 3, 4],
?[5, 6, 7, 8],
?[9,10,11,12]
]
输出: [1,2,3,4,8,12,11,10,9,5,6,7]
class Solution {
public:
vector<int> spiralOrder(vector<vector<int>>& matrix) {
/*
思路:
同59 螺旋矩阵 2
*/
vector<int> res;
if(matrix.empty()) return res;
int up = 0;
int down = matrix.size() - 1; //col
int left = 0;
int right = matrix[0].size() - 1; //row
while(up <= down && left <= right){
//left --> right
for(int i = left; i <= right; i++){
res.push_back(matrix[up][i]); //添加到res
}
//下一步要走的方向,依次类推
if(++up > down) break;
//up --> down
for(int i = up; i <= down; i++){
res.push_back(matrix[i][right]);
}
if(--right < left) break;
//right --> left
for(int i = right; i >= left; i--){
res.push_back(matrix[down][i]);
}
if(--down < up) break;
//down --> up
for(int i = down; i >= up; i--){
res.push_back(matrix[i][left]);
}
if(++left > right) break;
}
return res;
}
};
原文:https://www.cnblogs.com/fightingcode/p/11047082.html