• LeetCode 329 矩阵中最长增长路径


    LeetCode 329 矩阵中最长增长路径

    取自官方题解

    • 记忆化深度遍历
    class Solution {
        //方向矩阵: 上、下、左、右
        public int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        public int rows, columns;
    
        public int longestIncreasingPath(int[][] matrix) {
            //边界条件
            if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
                return 0;
            }
            rows = matrix.length;
            columns = matrix[0].length;
            //记忆数组
            int[][] memo = new int[rows][columns];
            int ans = 0;
            for (int i = 0; i < rows; ++i) {
                for (int j = 0; j < columns; ++j) {
                    ans = Math.max(ans, dfs(matrix, i, j, memo));
                }
            }
            return ans;
        }
        //递归求解子问题
        public int dfs(int[][] matrix, int row, int column, int[][] memo) {
            //取子问题解: 由(row, column)点出发的最长递增路径
            if (memo[row][column] != 0) {
                return memo[row][column];
            }
            //memo[row][column]至少为1,此时该路径下只有点(row, column)
            ++memo[row][column];
            //由点(row, column)向四个方向上搜索
            for (int[] dir : dirs) {
                int newRow = row + dir[0], newColumn = column + dir[1];
                if (//边界控制
                    newRow >= 0 && 
                    newRow < rows && 
                    newColumn >= 0 && 
                    newColumn < columns && 
                    //满足路径增长条件(增加)
                    matrix[newRow][newColumn] > matrix[row][column]) {
                    //求解子问题: (row, column)出发的最长递增路径,并存储
                    memo[row][column] = Math.max(memo[row][column], dfs(matrix, newRow, newColumn, memo) + 1);
                }
            }
            return memo[row][column];
        }
    }
    
    • 有向无环图拓扑排序求最长路径
    class Solution {
        public int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        public int rows, columns;
    
        public int longestIncreasingPath(int[][] matrix) {
            if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
                return 0;
            }
            rows = matrix.length;
            columns = matrix[0].length;
            //出度矩阵(小值点->大值点),用于存储生成的有向图中每个节点的出度数量
            int[][] outdegrees = new int[rows][columns];
            for (int i = 0; i < rows; ++i) {
                for (int j = 0; j < columns; ++j) {
                    for (int[] dir : dirs) {
                        int newRow = i + dir[0], newColumn = j + dir[1];
                        if (newRow >= 0 && newRow < rows && newColumn >= 0 && newColumn < columns && 
                        matrix[newRow][newColumn] > matrix[i][j]) {
                            ++outdegrees[i][j];
                        }
                    }
                }
            }
            Queue<int[]> queue = new LinkedList<int[]>();
            for (int i = 0; i < rows; ++i) {
                for (int j = 0; j < columns; ++j) {
                    //寻找图中所有无出度的节点(每条增长路径的终点)
                    if (outdegrees[i][j] == 0) {
                        queue.offer(new int[]{i, j});
                    }
                }
            }
            int ans = 0;
            while (!queue.isEmpty()) {
                ++ans;
                int size = queue.size();
                //遍历并删除当前图中每个无出度节点,该过程中同时添加新产生的无出度节点
                for (int i = 0; i < size; ++i) {
                    int[] point = queue.poll();
                    int row = point[0], column = point[1];
                    for (int[] dir : dirs) {
                        int newRow = row + dir[0], newColumn = column + dir[1];
                        if (newRow >= 0 && newRow < rows && newColumn >= 0 && newColumn < columns && 
                        matrix[newRow][newColumn] < matrix[row][column]) {
                            --outdegrees[newRow][newColumn];
                            if (outdegrees[newRow][newColumn] == 0) {
                                queue.offer(new int[]{newRow, newColumn});
                            }
                        }
                    }
                }
            }
            return ans;
        }
    }
    
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  • 原文地址:https://www.cnblogs.com/CodeSPA/p/13389716.html
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