329. Longest Increasing Path in a Matrix
难度Hard
刷题内容
原题连接
- https://leetcode.com/problems/longest-increasing-path-in-a-matrix/
内容描述
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:
Input: nums =
[
[9,9,4],
[6,6,8],
[2,1,1]
]
Output: 4
Explanation: The longest increasing path is [1, 2, 6, 9].
Example 2:
Input: nums =
[
[3,4,5],
[3,2,6],
[2,2,1]
]
Output: 4
Explanation: The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.
思路 **- 时间复杂度: O(n * m)*-空间复杂度: O(n * m)***
如果直接用DFS去做，时间复杂度为O(n^4)，我们就需要对算法进行优化，定义一个二维数组用记忆化的方法去记录进行剪枝
class Solution {
public:
int** dp;
int len1,len2,ans;
int dfs(vector<vector<int> >&matrix,int i,int j)
{
if(dp[i][j] != -1)
return dp[i][j];
//cout << i << j << endl;
if(i && matrix[i][j] < matrix[i - 1][j])
dp[i][j] = max(dp[i][j],dfs(matrix,i - 1,j) + 1);
if(j < len2 - 1 && matrix[i][j] < matrix[i][j + 1])
dp[i][j] = max(dp[i][j],dfs(matrix,i,j + 1) + 1);
if(i < len1 - 1 && matrix[i][j] < matrix[i + 1][j])
dp[i][j] = max(dp[i][j],dfs(matrix,i + 1,j) + 1);
if(j && matrix[i][j] < matrix[i][j - 1])
dp[i][j] = max(dp[i][j],dfs(matrix,i,j - 1) + 1);
//cout << dp[i][j];
if(dp[i][j] == -1)
dp[i][j] = 1;
ans = max(ans,dp[i][j]);
return dp[i][j];
}
int longestIncreasingPath(vector<vector<int>>& matrix) {
len1 = matrix.size();
if(!len1)
return 0;
len2 = matrix[0].size();
ans = 0;
dp = new int*[len1];
for(int i = 0; i < len1; ++i){
dp[i] = new int[len2];
for(int j=0;j<len2;j++)
dp[i][j]= -1;
}
//cout << 1;
for(int i = 0;i < matrix.size();++i)
for(int j = 0;j < matrix[i].size();++j)
if(dp[i][j] == -1)
{
//cout << 1;
dfs(matrix,i,j);
}
reverse(matrix.begin(),matrix.end());
for(int i = 0;i < matrix.size();++i)
reverse(matrix[i].begin(),matrix[i].end());
for(int i = 0;i < len1;++i)
for(int j = 0;j < len2;++j)
dp[i][j] = -1;
for(int i = 0;i < matrix.size();++i)
for(int j = 0;j < matrix[i].size();++j)
if(dp[i][j] == -1)
dfs(matrix,i,j);
return ans;
}
};
