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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.
这是一道典型的拓扑排序问题。
class Solution { public: struct node { int x; int y; }; queue<node> dict; int checkru(vector<vector<int>>& matrix,int a,int b,int m,int n) { int r=0; if(a>0) { if(matrix[a][b]>matrix[a-1][b]) r++; } if(a<m-1) { if(matrix[a][b]>matrix[a+1][b]) r++; } if(b>0) { if(matrix[a][b]>matrix[a][b-1]) r++; } if(b<n-1) { if(matrix[a][b]>matrix[a][b+1]) r++; } return r; } int longestIncreasingPath(vector<vector<int>>& matrix) { int m = matrix.size(); int n; int i,j; vector<vector<int>> ru; vector<vector<int>> maxpath; int max=1; if(m==0) return 0; n = matrix[0].size(); if(n==0) return 0; for(i=0;i<m;i++) { vector<int> k; vector<int> path; for(j=0;j<n;j++) { node nd; int uu; nd.x=i; nd.y=j; uu = checkru(matrix,i,j,m,n); k.push_back(uu); path.push_back(1); if(uu==0) dict.push(nd); } ru.push_back(k); maxpath.push_back(path); } while(!dict.empty()) { node nd = dict.front(); node kk; int a = nd.x; int b = nd.y; if(a>0) { if(matrix[a][b]<matrix[a-1][b]) { kk.x=a-1;kk.y=b; if(maxpath[a][b]+1>maxpath[a-1][b]) maxpath[a-1][b] = maxpath[a][b]+1; ru[a-1][b]--; if(ru[a-1][b]==0) dict.push(kk); if(maxpath[a-1][b]>max) max = maxpath[a-1][b]; } } if(b>0) { if(matrix[a][b]<matrix[a][b-1]) { kk.x=a;kk.y=b-1; if(maxpath[a][b]+1>maxpath[a][b-1]) maxpath[a][b-1] = maxpath[a][b]+1; ru[a][b-1]--; if(ru[a][b-1]==0) dict.push(kk); if(maxpath[a][b-1]>max) max = maxpath[a][b-1]; } } if(a<m-1) { if(matrix[a][b]<matrix[a+1][b]) { kk.x=a+1;kk.y=b; if(maxpath[a][b]+1>maxpath[a+1][b]) maxpath[a+1][b] = maxpath[a][b]+1; ru[a+1][b]--; if(ru[a+1][b]==0) dict.push(kk); if(maxpath[a+1][b]>max) max = maxpath[a+1][b]; } } if(b<n-1) { if(matrix[a][b]<matrix[a][b+1]) { kk.x=a;kk.y=b+1; if(maxpath[a][b]+1>maxpath[a][b+1]) maxpath[a][b+1] = maxpath[a][b]+1; ru[a][b+1]--; if(ru[a][b+1]==0) dict.push(kk); if(maxpath[a][b+1]>max) max = maxpath[a][b+1]; } } dict.pop(); } return max; } };
leetcode 329. Longest Increasing Path in a Matrix
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原文地址：http://www.cnblogs.com/julyfrost/p/5445780.html
