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带你快速刷完67道剑指offer

No19、顺时针打印矩阵

牛客网原题链接

题目描述

输入一个矩阵，按照从外向里以顺时针的顺序依次打印出每一个数字，例如，如果输入如下4 X 4矩阵： 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 则依次打印出数字1,2,3,4,8,12,16,15,14,13,9,5,6,7,11,10.

示例1

输入

[[1,2],[3,4]]

返回值

[1,2,4,3]

1、有点难，在力扣上写了好久

主要就是分析清楚上下左右的情况

执行用时：0 ms, 在所有 C++ 提交中击败了100.00%的用户

内存消耗：6.7 MB, 在所有 C++ 提交中击败了100.00%的用户

vector<int> spiralOrder(vector<vector<int>>& matrix) {
    if (matrix.size()==0) return vector<int>();
    if (matrix.size() == 1) return matrix[0];
    int row = matrix.size(), col = matrix[0].size();
    int left = 0, right = 0, top = 0, bottom = 0;
    vector<int> result;
    while (left + right < col && top + bottom < row) {
        
        for (int i = left; i < col - left - right + left; ++i) {
            //cout << matrix[top][i];
            result.push_back(matrix[top][i]);
        }

        top++;
        //cout << " top " <<top<<bottom<< endl;
        if (top + bottom == row) break;


        for (int i = top; i < row - top - bottom + top; ++i) {
            //cout << matrix[i][col - right - 1];
            result.push_back(matrix[i][col - right - 1]);
        }       
        right++;
        //cout << "right"<<left<<right<<endl;
        if (left + right == col) break;


        for (int i = col-right-1; i >= left ; --i) {
            //cout << matrix[row - bottom - 1][i];
            result.push_back(matrix[row - bottom - 1][i]);
        }
        bottom++;
        //cout << " bottom " << top << bottom << endl;
        if (top + bottom == row) break;
        

        for (int i = row-bottom-1; i >= top; --i) {
            //cout << matrix[i][left];
            result.push_back(matrix[i][left]);
        }
        left++;
        //cout << "left" << left << right << endl;
    }
    return result;
}

2、新的写法，这种其实更好理解

执行用时：24 ms, 在所有 C++ 提交中击败了56.85%的用户

内存消耗：10 MB, 在所有 C++ 提交中击败了100.00%的用户

vector<int> spiralOrder(vector<vector<int>>& matrix) {
        vector <int> res;
        if(matrix.empty()) return res;
        int rl = 0, rh = matrix.size()-1; //记录待打印的矩阵上下边缘
        int cl = 0, ch = matrix[0].size()-1; //记录待打印的矩阵左右边缘
        while(1){
            for(int i=cl; i<=ch; i++) res.push_back(matrix[rl][i]);//从左往右
            if(++rl > rh) break; //若超出边界，退出
            for(int i=rl; i<=rh; i++) res.push_back(matrix[i][ch]);//从上往下
            if(--ch < cl) break;
            for(int i=ch; i>=cl; i--) res.push_back(matrix[rh][i]);//从右往左
            if(--rh < rl) break;
            for(int i=rh; i>=rl; i--) res.push_back(matrix[i][cl]);//从下往上
            if(++cl > ch) break;
        }
        return res;
    }

3、改进一下第二种写法，快上不少

执行用时：12 ms, 在所有 C++ 提交中击败了98.41%的用户

内存消耗：10.3 MB, 在所有 C++ 提交中击败了100.00%的用户

vector<int> spiralOrder(vector<vector<int>>& matrix) {
    vector <int> res;
    if (matrix.empty()) return res;
    int top = 0, bottom = matrix.size() - 1; //记录待打印的矩阵上下边缘
    int left = 0, right = matrix[0].size() - 1; //记录待打印的矩阵左右边缘
    while (1) {
        for (int i = left; i <= right; ++i) res.push_back(matrix[top][i]);//从左往右
        if (++top > bottom) break; //若超出边界，退出

        for (int i = top; i <= bottom; ++i) res.push_back(matrix[i][right]);//从上往下
        if (--right < left) break;

        for (int i = right; i >= left; --i) res.push_back(matrix[bottom][i]);//从右往左
        if (--bottom < top) break;

        for (int i = bottom; i >= top; --i) res.push_back(matrix[i][left]);//从下往上
        if (++left > right) break;
    }
    return res;
}

二刷：

1、最快的做法，注意中间的判断条件不可少

运行时间：3ms 占用内存：496k

 vector<int> printMatrix(vector<vector<int> > matrix) {

if (matrix.size() == 0 || matrix[0].size() == 0) return vector<int>();
    int left = 0, right = matrix[0].size() - 1, top = 0, bottom = matrix.size() - 1;
    vector<int> result;
    while (left <= right && top <= bottom) {
        for (int i = left; i <= right; ++i)
        {
            //cout << matrix[top][i] << " ";
            result.push_back(matrix[top][i]);

        }
        if (++top > bottom) break;
        for (int i = top; i <= bottom; ++i)
        {
            //cout << matrix[i][right] << " ";
            result.push_back(matrix[i][right]);

        }
        if (--right < left) break;
        for (int i = right ; i >= left; --i) {
            //cout << matrix[bottom][i] << " ";
            result.push_back(matrix[bottom][i]);
        }
        if (--bottom < top) break;
        for (int i = bottom; i >= top; --i) {
            //cout << matrix[i][left] << " ";
            result.push_back(matrix[i][left]);
        }
        if (++left > right) break;
    }

    return result;
    }