力扣hot100——矩阵

发布于:2024-12-20 ⋅ 阅读:(172) ⋅ 点赞:(0)

73. 矩阵置零

class Solution {
public:
    void setZeroes(vector<vector<int>>& a) {
        int n = a.size(), m = a[0].size();
        vector<int> r(n + 10, 0);
        vector<int> c(m + 10, 0);
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (!a[i][j]) {
                    r[i] = 1;
                    c[j] = 1;
                }
            }
        }

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (r[i] || c[j]) a[i][j] = 0;
            }
        }
    }
};

模拟

 54. 螺旋矩阵

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& a) {
        int n = a.size(), m = a[0].size();
        int x = 0, y = 0;
        int sum = m * n;

        int sx = n - 1, sy = m;
        int dx = 1, dy = 1;
        vector<int> ans;
        while (sum) {
            for (int i = 1; i <= sy; i++) {
                ans.push_back(a[x][y]);
                y += dy;
                sum--;
            }
            dy *= -1;
            sy--;
            x += dx;
            y += dy;
            for (int i = 1; i <= sx; i++) {
                ans.push_back(a[x][y]);
                x += dx;
                sum--;
            }
            dx *= -1;
            sx--;
            y += dy;
            x += dx;
        }
        return ans;
    }
};

套路题,模拟

 48. 旋转图像

class Solution {
public:
    void rotate(vector<vector<int>>& a) {
        int n = a.size(), m = a[0].size();
        vector<vector<int>> ans(n, vector<int>(m, 0));
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                ans[j][n - i - 1] = a[i][j];
            }
        }
        swap(a, ans);
    }
};

套路题,模拟

240. 搜索二维矩阵 II

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& a, int target) {
        int n = a.size(), m = a[0].size();
        int x = 0, y = m - 1;
        while (x < n && y >= 0) {
            if (a[x][y] == target) return true;
            if (a[x][y] > target) y--;
            else x++;
        }
        return false;
    }
};

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