上岸算法 I LeetCode Weekly Contest 224解题报告

No.1 可以形成最大正方形的矩形数目

解题思路

维护最大正方形的边长并计数即可。

代码展示

class Solution {
    public int countGoodRectangles(int[][] rectangles) {
        int maxLen = 0, count = 0;
        for (int[] rec : rectangles) {
            int len = Math.min(rec[0], rec[1]);
            if (len > maxLen) {
                maxLen = len;
                count = 0;
            }
            if (len == maxLen) {
                count++;
            }
        }
        return count;
    }
}

No.2 同积元组

解题思路

用 Map 记录每一种乘积的因数列表。

代码展示

class Solution {
    public int tupleSameProduct(int[] nums) {
        Arrays.sort(nums);
        // map[i] 表示乘积为 i 的较小的因数的列表
        // 比如 a * b == i 则 map[i].add(min(a, b))
        Map<Integer, List<Integer>> map = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                int a = nums[i], b = nums[j];
                int p = a * b;
                if (!map.containsKey(p)) {
                    map.put(p, new ArrayList<>());
                }
                map.get(p).add(a); // a < b
            }
        }
        // 有 len 对乘积为 i
        // 它们能构成 4 * len * (len - 1) 个元组
        int count = 0;
        for (var entry : map.entrySet()) {
            int len = entry.getValue().size();
            count += 4 * len * (len - 1);
        }
        return count;
    }
}

No.3 重新排列后的最大子矩阵

解题思路

枚举最大子矩阵的起始行，然后排序、统计即可。

代码展示

class Solution {
    public int largestSubmatrix(int[][] matrix) {
        Column[] cols = new Column[matrix[0].length];
        for (int i = 0; i < matrix[0].length; i++) {
            int[] col = new int[matrix.length];
            for (int j = 0; j < matrix.length; j++) {
                col[j] = matrix[j][i];
            }
            cols[i] = new Column(col);
        }
        int res = 0;
        // 枚举最大子矩阵的起始行 i, 然后排序
        for (int i = 0; i < matrix.length; i++) {
            final int idx = i;
            Arrays.sort(cols, (a, b) -> b.cont1[idx] - a.cont1[idx]);
            if (res < cols[0].cont1[idx] * cols.length) { // 一个优化点
                res = Math.max(res, count(cols, idx));
            }
        }
        return res;
    }

    // 统计当前局面下的最大子矩阵
    private int count(Column[] cols, int startRow) {
        int res = 0;
        for (int endRow = startRow, endCol = cols.length - 1; endRow < cols[0].raw.length; endRow++) {
            while (endCol >= 0 && cols[endCol].raw[endRow] == 0) {
                endCol--;
            }
            res = Math.max(res, (endRow - startRow + 1) * (endCol + 1));
        }
        return res;
    }

    // Column 类表示一列
    // raw 表示这一列的原始值
    // cont1[i] 表示 raw[i] 后有多少个连续的 1
    class Column {
        int[] raw;
        int[] cont1;

        Column(int[] col) {
            raw = col;
            cont1 = new int[col.length];
            cont1[col.length - 1] = col[col.length - 1];
            for (int i = col.length - 2; i >= 0; i--) {
                cont1[i] = col[i];
                if (col[i] == 1) {
                    cont1[i] += cont1[i + 1];
                }
            }
        }
    }
}

No.4 猫和老鼠II

解题思路

动态规划（记忆化搜索）

虽然代码量比较大，但是逻辑并不复杂。本质上仍然是枚举猫和老鼠的下一步怎么走，老鼠想要赢就要保证它走下一步之后，猫无论怎么走都赢不了。然后使用记忆化以提速。

代码展示

class Solution {
    int n, m;
    char[][] grid;
    int[][][][][] catMemo;   // 记忆化
    int[][][][][] mouseMemo; // 记忆化
    int catJump, mouseJump;

    final int[][] dirs = new int[][] { { -1, 0 }, { 1, 0 }, { 0, 1 }, { 0, -1 } };
    final int MaxRound = 100; // 8x8 的地图远不需要 1000 步

    public boolean valid(int i, int j) {
        return i >= 0 && i < n && j >= 0 && j < m && grid[i][j] != '#';
    }

    // 已经走了 r 步，老鼠在 (mx, my) 猫在 (cx, cy)
    // 猫是否必胜
    public int cat(int r, int mx, int my, int cx, int cy) {
        if (catMemo[r][mx][my][cx][cy] == -1) {
            if (mx == cx && my == cy) {
                return catMemo[r][mx][my][cx][cy] = 1;
            }
            if (grid[mx][my] == 'F') {
                return catMemo[r][mx][my][cx][cy] = 0;
            }
            if (grid[cx][cy] == 'F') {
                return catMemo[r][mx][my][cx][cy] = 1;
            }
            for (int[] d : dirs) {
                for (int jump = 0; jump <= catJump; jump++) {
                    int x = cx + d[0] * jump;
                    int y = cy + d[1] * jump;
                    if (!valid(x, y)) {
                        break;
                    }
                    if (mouse(r + 1, mx, my, x, y) == 0) {
                        return catMemo[r][mx][my][cx][cy] = 1;
                    }
                }
            }
            catMemo[r][mx][my][cx][cy] = 0;
        }
        return catMemo[r][mx][my][cx][cy];
    }

    // 已经走了 r 步，老鼠在 (mx, my) 猫在 (cx, cy)
    // 老鼠是否必胜
    public int mouse(int r, int mx, int my, int cx, int cy) {
        if (r >= MaxRound) {
            return 0;
        }
        if (mouseMemo[r][mx][my][cx][cy] == -1) {
            if (mx == cx && my == cy) {
                return mouseMemo[r][mx][my][cx][cy] = 0;
            }
            if (grid[mx][my] == 'F') {
                return mouseMemo[r][mx][my][cx][cy] = 1;
            }
            if (grid[cx][cy] == 'F') {
                return mouseMemo[r][mx][my][cx][cy] = 0;
            }
            for (int[] d : dirs) {
                for (int jump = 0; jump <= mouseJump; jump++) {
                    int x = mx + d[0] * jump;
                    int y = my + d[1] * jump;
                    if (!valid(x, y)) {
                        break;
                    }
                    if (cat(r, x, y, cx, cy) == 0) {
                        return mouseMemo[r][mx][my][cx][cy] = 1;
                    }
                }
            }
            mouseMemo[r][mx][my][cx][cy] = 0;
        }
        return mouseMemo[r][mx][my][cx][cy];
    }

    public boolean canMouseWin(String[] grid, int catJump, int mouseJump) {
        n = grid.length;
        m = grid[0].length();
        this.grid = new char[n][m];
        int mx = 0, my = 0, cx = 0, cy = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                this.grid[i][j] = grid[i].charAt(j);
                if (this.grid[i][j] == 'C') {
                    cx = i;
                    cy = j;
                }
                if (this.grid[i][j] == 'M') {
                    mx = i;
                    my = j;
                }
            }
        }
        this.catJump = catJump;
        this.mouseJump = mouseJump;
        catMemo = new int[MaxRound][n][m][n][m];
        mouseMemo = new int[MaxRound][n][m][n][m];

        // 初值 -1 表示未计算过
        for (int r = 0; r < MaxRound; r++) {
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    for (int t = 0; t < n; t++) {
                        for (int k = 0; k < m; k++) {
                            catMemo[r][i][j][t][k] = -1;
                            mouseMemo[r][i][j][t][k] = -1;
                        }
                    }
                }
            }
        }
        return mouse(0, mx, my, cx, cy) == 1;
    }
}