329. Longest Increasing Path in a Matrix

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;

}
};