?LeetCode刷題實戰(zhàn)329：矩陣中的最長遞增路徑
Given an m x n integers matrix, return the length of the longest increasing path in matrix.
From each cell, you can either move in four directions: left, right, up, or down. You may not move diagonally or move outside the boundary (i.e., wrap-around is not allowed).
示例
解題
class Solution {
public int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
public int rows, columns;
public int longestIncreasingPath(int[][] matrix) {
if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
return 0;
}
rows = matrix.length;
columns = matrix[0].length;
int[][] memo = new int[rows][columns];
int ans = 0;
for (int i = 0; i < rows; ++i) {
for (int j = 0; j < columns; ++j) {
ans = Math.max(ans, dfs(matrix, i, j, memo));
}
}
return ans;
}
public int dfs(int[][] matrix, int row, int column, int[][] memo) {
if (memo[row][column] != 0) {
return memo[row][column];
}
++memo[row][column];
for (int[] dir : dirs) {
int newRow = row + dir[0], newColumn = column + dir[1];
if (newRow >= 0 && newRow < rows && newColumn >= 0 && newColumn < columns && matrix[newRow][newColumn] > matrix[row][column]) {
memo[row][column] = Math.max(memo[row][column], dfs(matrix, newRow, newColumn, memo) + 1);
}
}
return memo[row][column];
}
}
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