329. Longest Increasing Path in a Matrix

最长递增路径算法

最新推荐文章于 2022-05-19 10:30:09 发布

原创最新推荐文章于 2022-05-19 10:30:09 发布 · 296 阅读

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#back tracking

本文介绍了一种寻找矩阵中最长递增路径的算法。通过回溯法结合动态规划避免重复计算，实现高效求解。文章提供了详细的代码示例及解析。

Problem :

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.

Solution

如果只是用单纯的backtracking , 会做很多重复计算，所以关键是要想到用一个二维数组来储存从当前点出发的最长递增序列是多长

class Solution {
    int helper( int row , int col, const vector<vector<int>>& matrix, vector<vector<int>>& longestStartHere, int preVal){
        if(row < 0 || row >= matrix.size() || col < 0 || col >= matrix[0].size()) return 0;
        
        if( matrix[row][col] > preVal) {
            if(longestStartHere[row][col] != 0) return longestStartHere[row][col] ;
            int rst = 0 ;
             rst = max(rst,  helper( row + 1, col, matrix, longestStartHere, matrix[row][col]) + 1 );
             rst = max(rst,  helper( row - 1, col, matrix, longestStartHere, matrix[row][col]) + 1 );
             rst = max(rst,  helper( row, col + 1, matrix, longestStartHere, matrix[row][col]) + 1 );
             rst = max(rst,   helper( row, col - 1, matrix, longestStartHere, matrix[row][col]) + 1 );
            
            longestStartHere[row][col] = rst;
            return longestStartHere[row][col];
        }
        
        return 0;
    }
    
public:
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        if(matrix.empty() || matrix[0].empty()) return 0;
        
        const int M = matrix.size(), N = matrix[0].size();
        vector<vector<int>> longestStartHere (M , vector<int>(N, 0));
        int rst = 0;
        
        for( int i = 0; i < M; i++ ){
            for( int j = 0; j < N; j++){
                int preVal = matrix[i][j] - 1;
                rst = max( rst, helper(i, j, matrix, longestStartHere, INT_MIN));
            }
        }
        return rst;
        
    }
};