矩阵中最长连续递增子序列

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.

int x_1[4] = { 1, -1, 0, 0 };
int y_1[4] = { 0, 0, -1, 1 };

int dfs(vector<vector<int>>& matrix,int rows,int cols,
	int row, int col, vector<vector<int>>&dp)
{
	if (dp[row][col])
		return dp[row][col];
	int maxDepth = 1;
	for (int i = 0; i < 4; ++i)
	{
		int x = row + x_1[i];
		int y = col + y_1[i];
		if (x < 0 || y < 0 || x >= rows
			|| y >= cols || matrix[x][y] <= matrix[row][col])
			continue;
		int len = dfs(matrix, rows, cols, x, y, dp) + 1;
		if (maxDepth < len)
			maxDepth = len;
	}
	dp[row][col] = maxDepth;
	return maxDepth;
}

int longestIncreasingPath(vector<vector<int>>& matrix) 
{
	if (matrix.empty())
		return 0;
	int rows = matrix.size();
	int cols = matrix[0].size();
	vector<vector<int>> dp(rows, vector<int>(cols, 0));
	int maxDepth = 1;
	for (int i = 0; i < rows; ++i)
	{
		for (int j = 0; j < cols; ++j)
		{
			int tmp = dfs(matrix,rows,cols, i, j, dp);
			if (maxDepth < tmp)
				maxDepth = tmp;
		}
	}
	return maxDepth;
}


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转载自blog.csdn.net/yang20141109/article/details/70038183