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59. Spiral Matrix && Spiral Matrix II
Spiral Matrix
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; }};Spiral Matrix II
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
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