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;
    }
};