LeetCode每日一题【54.螺旋矩阵】

思路：模拟，初始化上下左右4个方向的边界，逐步收缩，注意要及时判断元素是否已经放满，否则会放入多余元素

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        int row = matrix.size();
        int col = matrix[0].size();
        vector<int> ans;
        int k = 1;
        int left = 0;
        int right = col-1;
        int top = 0;
        int bottom = row-1;
        while(k <= row*col){
            if(ans.size() == col * row)   break;
            for(int i = left;i<=right;++i,++k)  ans.push_back(matrix[top][i]);
            top++;

            if(ans.size() == col * row)   break;
            for(int i = top;i<=bottom;++i,++k)  ans.push_back(matrix[i][right]);
            right--;

            if(ans.size() == col * row)   break;
            for(int i = right;i>=left;--i,++k)  ans.push_back(matrix[bottom][i]);
            bottom--;
            
            if(ans.size() == col * row)   break;
            for(int i = bottom;i>=top;--i,++k)  ans.push_back(matrix[i][left]);
            left++;
        }
        return ans;
    }
};
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