Leetcode: Longest Increasing Path in a Matrix



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.

比较简单，DFS+DP。

class Solution {public:    int longestIncreasingPath(vector<vector<int>>& matrix) {        if (matrix.empty() || matrix[0].empty()) {            return 0;        }                int longestPath = 0;        int rows = matrix.size();        int cols = matrix[0].size();        vector<vector<int>> lip(rows, vector<int>(cols));        for (int i = 0; i < rows; ++i) {            for (int j = 0; j < cols; ++j) {                if (lip[i][j] == 0) {                    lip[i][j] = calculateIncreasingPath(matrix, lip, i, j);                }                longestPath = max(longestPath, lip[i][j]);            }        }                return longestPath;    }        int calculateIncreasingPath(vector<vector<int>>& matrix, vector<vector<int>>& lip, int i, int j) {        if (lip[i][j] != 0) {            return lip[i][j];        }                lip[i][j] = 1;        if (i > 0 && matrix[i][j] > matrix[i-1][j]) {            lip[i][j] = max(lip[i][j], calculateIncreasingPath(matrix, lip, i - 1, j) + 1);        }        if (i < matrix.size() - 1 && matrix[i][j] > matrix[i+1][j]) {            lip[i][j] = max(lip[i][j], calculateIncreasingPath(matrix, lip, i + 1, j) + 1);        }        if (j > 0 && matrix[i][j] > matrix[i][j-1]) {            lip[i][j] = max(lip[i][j], calculateIncreasingPath(matrix, lip, i, j - 1) + 1);        }        if (j < matrix[0].size() - 1 && matrix[i][j] > matrix[i][j+1]) {            lip[i][j] = max(lip[i][j], calculateIncreasingPath(matrix, lip, i, j + 1) + 1);        }                return lip[i][j];    }};