Question
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.
Code
public class Solution {
private int[] di = {1, 0, -1, 0};
private int[] dj = {0, 1, 0, -1};
private int max = 1;
public int longestIncreasingPath(int[][] matrix) {
if (matrix == null || matrix.length == 0) return 0;
int m = matrix.length, n = matrix[0].length;
int[][] dp = new int[m][n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
helper(dp, matrix, i, j);
}
}
return max;
}
public int helper(int[][] dp, int[][] matrix, int i, int j) {
if (dp[i][j] != 0) return dp[i][j];
int m = matrix.length, n = matrix[0].length;
dp[i][j] = 1;
for (int k = 0; k < 4; k++) {
int ni = i + di[k];
int nj = j + dj[k];
if (ni >= 0 && ni < m && nj >= 0 && nj < n) {
if (matrix[i][j] > matrix[ni][nj]) {
dp[i][j] = Math.max(dp[i][j], helper(dp, matrix, ni, nj) + 1);
}
}
}
max = Math.max(max, dp[i][j]);
return dp[i][j];
}
}
Solution
DFS + 動態(tài)規(guī)劃 + 記憶化搜索疙挺。
由于動態(tài)規(guī)劃的初始起點不好確定，因此使用動態(tài)規(guī)劃 + 記憶化搜索的方法完成虱黄。
dp[i][j]表示以matrix[i][j]為終點的最長遞增路徑的長度。通過DFS向四周搜索，如果matrix[四周i][四周j]小于matrix[i][j]，使用狀態(tài)轉(zhuǎn)移函數(shù)dp[i][j] = Math.max(dp[i][j], helper(dp, matrix, ni, nj) + 1); 最后返回dp中的最大值即可知态。
