p329

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LeetCode p329 Longest Increasing Path in a Matrix 题解

1.题目：

Given an integer matrix, find the length of the longest increasing path.

From each cell, you can either move to four directions: left, right, up or down. You may NOT move diagonally or move outside of the boundary (i.e. wrap-around is not allowed).

Example 1:

nums = [
[9,9,4],
[6,6,8],
[2,1,1]
]

Return 4
The longest increasing path is [1, 2, 6, 9].

Example 2:

nums = [
[3,4,5],
[3,2,6],
[2,2,1]
]

Return 4
The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.

题意：

给一个二维数组，求其内部可以连成的最长递增路径。
可与上下左右相连。

2.解题思路:

dfs 将每一点作为起点开始搜索，不过这样会超时。
所以设立缓存的数组dfs[][],进行储存。

3.代码

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public class Solution {
   public int longestIncreasingPath(int[][] matrix) {

	int ans = 0;
		int[][] dfs;
		if (matrix.length < 1)
			return 0;
		dfs = new int[matrix.length][matrix[0].length];
		for (int i = 0; i < matrix.length; i++) {
			for (int j = 0; j < matrix[0].length; j++) {
				dfs[i][j] = -1;
			}
		}
		for (int i = 0; i < matrix.length; i++) {
			for (int j = 0; j < matrix[0].length; j++) {
				ans = Math.max(
						ans,
						longestIncreasingPath(matrix, i, j, dfs,
								Integer.MIN_VALUE));
				// System.out.println(i + "  " + j + "   " + dfs[i][j]);
			}
		}
		return ans ;

	}

	public static int longestIncreasingPath(int[][] matrix, int i, int j,
			int[][] dfs, int a) {
		if (i < 0 || i >= matrix.length || j < 0 || j >= matrix[0].length
				|| a >= matrix[i][j]) {
			return 0;
		}
		if (dfs[i][j] > -1) {

			// System.out.println("----------"+dfs[i][j]+"i:"+i+"  "+j);

			return dfs[i][j];
		}
		int c = 0;
		// up

		// System.out.println(c);
		c = Math.max(c,
				longestIncreasingPath(matrix, i - 1, j, dfs, matrix[i][j]) + 1);

		// down

		c = Math.max(c,
				longestIncreasingPath(matrix, i + 1, j, dfs, matrix[i][j]) + 1);

		// left

		c = Math.max(c,
				longestIncreasingPath(matrix, i, j - 1, dfs, matrix[i][j]) + 1);

		// right

		c = Math.max(c,
				longestIncreasingPath(matrix, i, j + 1, dfs, matrix[i][j]) + 1);

		dfs[i][j] = c;
		return c;
			}
}

4.一些总结：