LeetCode 329. Longest Increasing Path in a Matrix

DFS

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.

解题思路：

找出以每个数作为起点所能走的最长路径，再输出其中最大的路径长度。考虑使用DFS，若当前某个数的四周有比其更大的数，则进入这个更大的数继续搜索。

为了简化过程，用一个另外的矩阵储存遍历图中的数所能到达的最大长度，若再次遍历到则不用重新计算。

代码如下：

class Solution {
public:
int longestIncreasingPath(vector<vector<int>>& matrix) {
if (matrix.size() == 0 || matrix[0].size() == 0) return 0;
this->matrix = matrix;
path.clear();

vector<int> inits;
for (int i = 0; i < matrix[0].size(); i++) {
inits.push_back(0);
}
for (int i = 0; i < matrix.size(); i++) {
path.push_back(inits);
}

int max = 1;

for (int i = 0; i < path.size(); i++) {
for (int j = 0; j < path[0].size(); j++) {
if (path[i][j] == 0) findPath(i, j);
if (path[i][j] > max)
max = path[i][j];
}
}

return max;
}

private:

void findPath(int x, int y) {
int tmpLength;

if (x < path.size()-1 && matrix[x][y] < matrix[x+1][y]) {
tmpLength = getLongest(x+1, y);
if (tmpLength > path[x][y]) path[x][y] = tmpLength;
}

if (x > 0 && matrix[x][y] < matrix[x-1][y]) {
tmpLe
4000
ngth = getLongest(x-1, y);
if (tmpLength > path[x][y]) path[x][y] = tmpLength;
}

if (y < path[0].size()-1 && matrix[x][y] < matrix[x][y+1]) {
tmpLength = getLongest(x, y+1);
if (tmpLength > path[x][y]) path[x][y] = tmpLength;
}

if (y > 0 && matrix[x][y] < matrix[x][y-1]) {
tmpLength = getLongest(x, y-1);
if (tmpLength > path[x][y]) path[x][y] = tmpLength;
}

path[x][y]++;

}

int getLongest(int x, int y) {
if (path[x][y] == 0)
findPath(x, y);
return path[x][y];
}

vector<vector<int> > path;
vector<vector<int> > matrix;

};

提交结果为：

137 / 137 test cases passed.

Status: Accepted

Runtime: 33 ms

Your runtime beats 75.21 % of cpp submissions.