329. Longest Increasing Path in a Matrix

    /** 329. Longest Increasing Path in a Matrix* 2016-7-9 by Mingyang* 其实这种类型的题目没有想象的那么难，你只需要好好的按照dfs的概念一点一点的走就好了* 记住，每个dfs不光是void的，也可以返回一个长度*/public static final int[][] dirs = { { 0, 1 }, { 1, 0 }, { 0, -1 }, { -1, 0 } };public int longestIncreasingPath(int[][] matrix) {if (matrix.length == 0)return 0;int m = matrix.length, n = matrix[0].length;int[][] cache = new int[m][n];int max = 1;for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {int len = dfs(matrix, i, j, m, n, cache);max = Math.max(max, len);}}return max;}public int dfs(int[][] matrix, int i, int j, int m, int n, int[][] cache) {if (cache[i][j] != 0)return cache[i][j];int max = 1;for (int[] dir : dirs) {int x = i + dir[0], y = j + dir[1];if (x < 0 || x >= m || y < 0 || y >= n || matrix[x][y] <= matrix[i][j])continue;//m,n is the boundryint len = 1 + dfs(matrix, x, y, m, n, cache);max = Math.max(max, len);}cache[i][j] = max;return max;}

转载于:https://www.cnblogs.com/zmyvszk/p/5657029.html

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