LeetCode:Longest Increasing Path in a Matrix
Longest Increasing Path in a Matrix
Total Accepted: 15060
Total Submissions: 46586
Difficulty: Hard
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.
Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.
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思路：
DFS.
java code:
public class Solution {
public int longestIncreasingPath(int[][] matrix) {
if(matrix == null || matrix.length == 0) return 0;
int rows = matrix.length;
int cols = matrix[0].length;
int[][] maxLen = new int[rows][cols];
int ans = 1;
for(int i=0;i<rows;i++) {
for(int j=0;j<cols;j++) {
ans = Math.max(ans, DFS(matrix, maxLen, i, j, rows, cols));
}
}
return ans;
}
// 自定义函数
private static final int[][] dirs = {{0,-1},{0,1},{-1,0},{1,0}}; // 方向
private int DFS(int[][] matrix, int[][] maxLen, int i, int j, int rows, int cols) {
if(maxLen[i][j] != 0) return maxLen[i][j];
int max = 1;
for(int[] dir : dirs) {
int x = i + dir[0], y = j + dir[1];
if(x < 0 || x >= rows || y < 0 || y >= cols) continue;
if(matrix[i][j] <= matrix[x][y]) continue;
max = Math.max(max, 1 + DFS(matrix, maxLen, x, y, rows, cols));
}
maxLen[i][j] = max;
return max;
}
}