题目描述
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.
分析解答
通过记忆化搜索实现动态规划，记忆化搜索是搜索和动态规划的结合，在搜索过程中记录已经求解完毕的状态，使得每个状态只需进行一次搜索。在本题中，cache[i][j]记录在(i,j)位置可以得到的最长上升路径，其值通过搜索相邻四个位置的最大路径值确定
1. 对每个点DFS，在4个方向找比当前小的数
2. 利用动态规划的思想得到每个点的路径最大值
3. 用cache数组存放距离，以后不用重复计算
参考代码
记忆化搜索
class Solution {
public:
const int dr[4] = { 1, 0, -1, 0 };
const int dc[4] = { 0, -1, 0, 1 };
int dfs(vector<vector<int>>& matrix, int x, int y, vector<vector<int>>& cache) {
// if calculated before, no need to do it again
if (cache[x][y]) return cache[x][y];
int res = 1;
for (int i=0; i<4; i++) {
int nx = x + dr[i];
int ny = y + dc[i];
// if out of bond OR current cell value larger than previous cell value.
if (nx < 0 || nx >= matrix.size() || ny < 0 || ny >= matrix[0].size() ||
matrix[nx][ny] >= matrix[x][y]) continue;
res = max(res, dfs(matrix, nx, ny, cache) + 1);
}
return cache[x][y] = res;
}
int longestIncreasingPath(vector<vector<int>>& matrix) {
if (matrix.empty() || matrix[0].empty()) return 0;
int m = matrix.size(), n = matrix[0].size();
vector<vector<int>> cache(m, vector<int>(n, 0));
int res = 0;
for (int i=0; i<m; i++) {
for (int j=0; j<n; j++) {
cache[i][j] = dfs(matrix, i, j, cache);
res = max(res, cache[i][j]);
}
}
return res;
}
};
参考资料
LeetCode 329. Longest Increasing Path in a Matrix
15ms Concise Java Solution
---------------------
作者：epsilon1
来源：CSDN
原文：https://blog.csdn.net/qq_20480611/article/details/52118059
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