Longest Increasing Path in a Matrix
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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.
题目的大意就是给定一个数组,找出数字按递增排列的最长路径的长度。
这道题可以利用深搜的办法,给定一个起始点,对这个点四个方向的点做一次判断,若相邻的点比当前点大,那么就递归调用一次DepthFirst函数,得到以该相邻点为起始点的最大路径长度,然后比较四个相邻点的最大路径长度的最大值,取最大值加到当前点的最大路径长度中。
为了降低复杂度,需要使用两个辅助数组,一个数组是记录每个点的对应于以这个点为起点的最大路径长度,这个数组记为count,另一个数组b则是记录该点是否已经访问过,在调用递归函数之前,要先判断该点是否已经访问过,如果已经访问过就不需要再次调用函数了。
另外还要用一个变量maximum记录当前的最大长度,这样就不用在最后还要进行寻找最大值的操作。而且在主函数中要用一个两重循环来确定每一个点都被访问过。
这个算法是一个深度优先算法,所以时间复杂度为O(M*N).
还有一点需要特别注意的是,在传递数组的参数时,一定要记得将变量定义为引用类型,否则在每次调用函数时数组都要重新复制一次,我就是因为没有注意到这个问题,导致我在LeetCode上提交时超时了,找了很久才找到原因。
以下为源程序:
class Solution {public: int maximum=1,n,m; int longestIncreasingPath(vector<vector<int>>& matrix) { if(matrix.size()==0) return 0; n=matrix.size(); m=matrix[0].size(); vector<vector<bool> > b(n,vector<bool>(m,false)); vector<vector<int> > count(n,vector<int>(m,1)); DepthFirst(matrix,count,b,0,0); for(int i=0;i<n;i++) for(int j=0;j<m;j++) if(!b[i][j]) { DepthFirst(matrix,count,b,i,j); } return maximum; } void DepthFirst(const vector<vector<int> >& matrix,vector<vector<int> >& count,vector<vector<bool> >& b,int i,int j) { int num=0; b[i][j]=true; if(i-1>=0&&matrix[i-1][j]>matrix[i][j]) { if(!b[i-1][j]) DepthFirst(matrix,count,b,i-1,j); if(count[i-1][j]>num) num=count[i-1][j]; } if(j-1>=0&&matrix[i][j-1]>matrix[i][j]) { if(!b[i][j-1]) DepthFirst(matrix,count,b,i,j-1); if(count[i][j-1]>num) num=count[i][j-1]; } if(i+1<n&&matrix[i+1][j]>matrix[i][j]) { if(!b[i+1][j]) DepthFirst(matrix,count,b,i+1,j); if(count[i+1][j]>num) num=count[i+1][j]; } if(j+1<m&&matrix[i][j+1]>matrix[i][j]) { if(!b[i][j+1]) DepthFirst(matrix,count,b,i,j+1); if(count[i][j+1]>num) num=count[i][j+1]; } count[i][j]+=num; if(count[i][j]>maximum) maximum=count[i][j]; }};
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