[Leetcode][第329题][JAVA][矩阵中的最长递增路径][DFS][拓扑排序]

【问题描述】[中等]

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【解答思路】

1. 记忆化深度优先搜索

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复杂度
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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;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;}public int dfs(int[][] matrix, int row, int column, int[][] memo) {if (memo[row][column] != 0) {return memo[row][column];}++memo[row][column];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]) {memo[row][column] = Math.max(memo[row][column], dfs(matrix, newRow, newColumn, memo) + 1);}}return memo[row][column];}
}

2. 拓扑排序

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复杂度
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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;int[][] outdegrees = new int[rows][columns];//初始化度 度递增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];}}}}//边界条件入度 相关点进行递减Queue<int[]> queue = new LinkedList<int[]>();for (int i = 0; i < rows; ++i) {for (int j = 0; j < columns; ++j) {if (outdegrees[i][j] == 0) {queue.offer(new int[]{i, j});}}}int ans = 0;while (!queue.isEmpty()) {++ans;int size = queue.size();for (int i = 0; i < size; ++i) {int[] cell = queue.poll();int row = cell[0], column = cell[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;}
}