[leetcode] 329. 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.

这道题是求矩阵中最长递增路径长度，题目难度为Hard。

​典型的DFS题目，遍历矩阵中所有位置的数字，以此位置为初始位置，分别向四个方向上深度优先遍历，这样既可得出最长路径长度，不过提交之后超时了。

我们发现，对所有位置进行深度优先遍历的时候，可能会有很多位置被重复遍历，如果记录下已经遍历过的位置的最长递增路径长度，这样在从其他位置深度优先遍历到此位置时就不用再继续遍历了，只需要返回记录的最长路径即可。这里用动态规划的思想拿空间换时间，提高了搜索效率。具体代码：

class Solution {
    int getPathLen(const vector>& matrix, vector>& maxLen, int r, int c) {
        int maxPathLen = 1;
        
        if(maxLen[r][c]) return maxLen[r][c];
        
        if(r>0 && matrix[r-1][c]>matrix[r][c]) 
            maxPathLen = max(maxPathLen, getPathLen(matrix, maxLen, r-1, c)+1);
        if(rmatrix[r][c]) 
            maxPathLen = max(maxPathLen, getPathLen(matrix, maxLen, r+1, c)+1);
        if(c>0 && matrix[r][c-1]>matrix[r][c])
            maxPathLen = max(maxPathLen, getPathLen(matrix, maxLen, r, c-1)+1);
        if(cmatrix[r][c])
            maxPathLen = max(maxPathLen, getPathLen(matrix, maxLen, r, c+1)+1);
        
        maxLen[r][c] = maxPathLen;
        
        return maxPathLen;
    }
public:
    int longestIncreasingPath(vector>& matrix) {
        if(matrix.empty() || matrix[0].empty()) return 0;
        int ret = 0, row = matrix.size(), col = matrix[0].size();
        vector> maxLen(row, vector(col, 0));
        for(int r=0; r