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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.
Analyse: backtracking.
Time Limit Exceeded Version.
1 class Solution { 2 public: 3 int longestIncreasingPath(vector<vector<int>>& matrix) { 4 int result; 5 if(matrix.empty() || matrix[0].empty()) return result; 6 7 int depth = 0; 8 vector<vector<bool> > visited(matrix.size(), vector<bool>(matrix[0].size(), false)); 9 for(int i = 0; i < matrix.size(); i++){ 10 for(int j = 0; j < matrix[i].size(); j++){ 11 helper(matrix, visited, i, j, result, depth, INT_MIN); 12 } 13 } 14 return result; 15 } 16 17 void helper(vector<vector<int> >& matrix, vector<vector<bool> > visited, int i, int j, int& result, int depth, int former){ 18 if(i >= matrix.size() || i < 0 || j >= matrix[i].size() || j < 0 || matrix[i][j] <= former || visited[i][j]) 19 return; 20 21 visited[i][j] = true; 22 depth++; 23 result = max(result, depth); 24 former = matrix[i][j]; 25 helper(matrix, visited, i + 1, j, result, depth, former); 26 helper(matrix, visited, i - 1, j, result, depth, former); 27 helper(matrix, visited, i, j + 1, result, depth, former); 28 helper(matrix, visited, i, j - 1, result, depth, former); 29 visited[i][j] = false; 30 } 31 };
Longest Increasing Path in a Matrix
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原文地址：http://www.cnblogs.com/amazingzoe/p/5226193.html
