标签：style blog http io ar for div sp log
Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.
For example, Given the following matrix:
[ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ]
You should return [1,2,3,6,9,8,7,4,5].
思路： 可参考剑指offer：题20
class Solution {
public:
vector<int> spiralOrder(vector<vector<int> > &matrix) {
vector<int> vec;
if(!matrix.size() || !matrix[0].size()) return vec;
int row = matrix.size(), col = matrix[0].size();
int rl = 0, rh = row-1, cl = 0, ch = col-1;
while(rh >= rl && ch >= cl) {
for(int c = cl; c <= ch; ++c) vec.push_back(matrix[rl][c]);
if(++rl > rh) break;
for(int r = rl; r <= rh; ++r) vec.push_back(matrix[r][ch]);
if(--ch < cl) break;
for(int c = ch; c >= cl; --c) vec.push_back(matrix[rh][c]);
--rh;
for(int r = rh; r >= rl; --r) vec.push_back(matrix[r][cl]);
++cl;
}
return vec;
}
};
Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.
For example, Given n = 3,
You should return the following matrix:
[ [ 1, 2, 3 ], [ 8, 9, 4 ], [ 7, 6, 5 ] ]
class Solution {
public:
vector<vector<int> > generateMatrix(int n) {
vector<vector<int> > vec(n, vector<int>(n));
int rl = 0, rh = n-1, cl = 0, ch = n-1, v = 1;
while(rh >= rl && ch >= cl) {
for(int c = cl; c <= ch; ++c) vec[rl][c] = v++;
if(++rl > rh) break;
for(int r = rl; r <= rh; ++r) vec[r][ch] = v++;
if(--ch < cl) break;
for(int c = ch; c >= cl; --c) vec[rh][c] = v++;
--rh;
for(int r = rh; r >= rl; --r) vec[r][cl] = v++;
++cl;
}
return vec;
}
};
59. Spiral Matrix && Spiral Matrix II
标签：style blog http io ar for div sp log
原文地址：http://www.cnblogs.com/liyangguang1988/p/3961165.html
