矩阵中的最长递增路径的长度
class Solution {
public:
//记录四个方向
int dirs[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
int n, m;
//深度优先搜索,返回最大单元格数
int dfs(vector > &matrix, vector > &dp, int i, int j) {
if(dp[i][j] != 0)
return dp[i][j];
dp[i][j]++;
for(int k = 0; k < 4; k++){
int nexti = i + dirs[k][0];
int nextj = j + dirs[k][1];
//判断条件
if (nexti >= 0 && nexti < n && nextj >= 0 && nextj < m && matrix[nexti][nextj] > matrix[i][j])
dp[i][j] = max(dp[i][j], dfs(matrix, dp, nexti, nextj) + 1);
}
return dp[i][j];
}
int solve(vector >& matrix) {
//矩阵不为空
if(matrix.size() == 0 || matrix[0].size() == 0)
return 0;
int res = 0;
n = matrix.size();
m = matrix[0].size();
//i,j处的单元格拥有的最长递增路径
vector > dp (n, vector (m));
for(int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
//更新最大值
res = max(res, dfs(matrix, dp, i, j));
return res;
}
};
public:
//记录四个方向
int dirs[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
int n, m;
//深度优先搜索,返回最大单元格数
int dfs(vector > &matrix, vector > &dp, int i, int j) {
if(dp[i][j] != 0)
return dp[i][j];
dp[i][j]++;
for(int k = 0; k < 4; k++){
int nexti = i + dirs[k][0];
int nextj = j + dirs[k][1];
//判断条件
if (nexti >= 0 && nexti < n && nextj >= 0 && nextj < m && matrix[nexti][nextj] > matrix[i][j])
dp[i][j] = max(dp[i][j], dfs(matrix, dp, nexti, nextj) + 1);
}
return dp[i][j];
}
int solve(vector >& matrix) {
//矩阵不为空
if(matrix.size() == 0 || matrix[0].size() == 0)
return 0;
int res = 0;
n = matrix.size();
m = matrix[0].size();
//i,j处的单元格拥有的最长递增路径
vector > dp (n, vector (m));
for(int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
//更新最大值
res = max(res, dfs(matrix, dp, i, j));
return res;
}
};
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