最长递增路径

剑指 Offer II 112. 最长递增路径

给定一个 m x n 整数矩阵 matrix ，找出其中 最长递增路径 的长度。

对于每个单元格，你可以往上，下，左，右四个方向移动。不能在 对角线 方向上移动或移动到 边界外（即不允许环绕）。

示例 1：

输入：matrix = [[9,9,4],[6,6,8],[2,1,1]]
输出：4 
解释：最长递增路径为 [1, 2, 6, 9]。

示例 2：

输入：matrix = [[3,4,5],[3,2,6],[2,2,1]]
输出：4 
解释：最长递增路径是 [3, 4, 5, 6]。注意不允许在对角线方向上移动。

示例 3：

输入：matrix = [[1]]
输出：1

提示：

m == matrix.length
n == matrix[i].length
1 <= m, n <= 200
0 <= matrix[i][j] <= 2³¹ - 1

注意：本题与主站 329 题相同： https://leetcode.cn/problems/longest-increasing-path-in-a-matrix/

原站题解

去查看

上次编辑到这里，代码来自缓存点击恢复默认模板

class Solution { public: int longestIncreasingPath(vector<vector<int>>& matrix) { } };

java 解法, 执行用时: 15 ms, 内存消耗: 41.8 MB, 提交时间: 2023-04-22 12:29:05

class Solution {
    public int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
    public int rows, columns;

    public int longestIncreasingPath(int[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return 0;
        }
        rows = matrix.length;
        columns = matrix[0].length;
        int[][] outdegrees = new int[rows][columns];
        for (int i = 0; i < rows; ++i) {
            for (int j = 0; j < columns; ++j) {
                for (int[] dir : dirs) {
                    int newRow = i + dir[0], newColumn = j + dir[1];
                    if (newRow >= 0 && newRow < rows && newColumn >= 0 && newColumn < columns && matrix[newRow][newColumn] > matrix[i][j]) {
                        ++outdegrees[i][j];
                    }
                }
            }
        }
        Queue<int[]> queue = new LinkedList<int[]>();
        for (int i = 0; i < rows; ++i) {
            for (int j = 0; j < columns; ++j) {
                if (outdegrees[i][j] == 0) {
                    queue.offer(new int[]{i, j});
                }
            }
        }
        int ans = 0;
        while (!queue.isEmpty()) {
            ++ans;
            int size = queue.size();
            for (int i = 0; i < size; ++i) {
                int[] cell = queue.poll();
                int row = cell[0], column = cell[1];
                for (int[] dir : dirs) {
                    int newRow = row + dir[0], newColumn = column + dir[1];
                    if (newRow >= 0 && newRow < rows && newColumn >= 0 && newColumn < columns && matrix[newRow][newColumn] < matrix[row][column]) {
                        --outdegrees[newRow][newColumn];
                        if (outdegrees[newRow][newColumn] == 0) {
                            queue.offer(new int[]{newRow, newColumn});
                        }
                    }
                }
            }
        }
        return ans;
    }
}

python3 解法, 执行用时: 280 ms, 内存消耗: 16 MB, 提交时间: 2023-04-22 12:28:42

# 拓扑排序
class Solution:

    DIRS = [(-1, 0), (1, 0), (0, -1), (0, 1)]

    def longestIncreasingPath(self, matrix: List[List[int]]) -> int:
        if not matrix:
            return 0
        
        rows, columns = len(matrix), len(matrix[0])
        outdegrees = [[0] * columns for _ in range(rows)]
        queue = collections.deque()
        for i in range(rows):
            for j in range(columns):
                for dx, dy in Solution.DIRS:
                    newRow, newColumn = i + dx, j + dy
                    if 0 <= newRow < rows and 0 <= newColumn < columns and matrix[newRow][newColumn] > matrix[i][j]:
                        outdegrees[i][j] += 1
                if outdegrees[i][j] == 0:
                    queue.append((i, j))

        ans = 0
        while queue:
            ans += 1
            size = len(queue)
            for _ in range(size):
                row, column = queue.popleft()
                for dx, dy in Solution.DIRS:
                    newRow, newColumn = row + dx, column + dy
                    if 0 <= newRow < rows and 0 <= newColumn < columns and matrix[newRow][newColumn] < matrix[row][column]:
                        outdegrees[newRow][newColumn] -= 1
                        if outdegrees[newRow][newColumn] == 0:
                            queue.append((newRow, newColumn))
        
        return ans

golang 解法, 执行用时: 36 ms, 内存消耗: 7.3 MB, 提交时间: 2023-04-22 12:28:14

var (
    dirs = [][]int{[]int{-1, 0}, []int{1, 0}, []int{0, -1}, []int{0, 1}}
    rows, columns int
)

func longestIncreasingPath(matrix [][]int) int {
    if len(matrix) == 0 || len(matrix[0]) == 0 {
        return 0
    }
    rows, columns = len(matrix), len(matrix[0])
    outdegrees := make([][]int, rows)
    for i := 0; i < rows; i++ {
        outdegrees[i] = make([]int, columns)
    }
    for i := 0; i < rows; i++ {
        for j := 0; j < columns; j++ {
            for _, dir := range dirs {
                newRow, newColumn := i + dir[0], j + dir[1]
                if newRow >= 0 && newRow < rows && newColumn >= 0 && newColumn < columns && matrix[newRow][newColumn] > matrix[i][j] {
                    outdegrees[i][j]++
                }
            }
        }
    }

    queue := [][]int{}
    for i := 0; i < rows; i++ {
        for j := 0; j < columns; j++ {
            if outdegrees[i][j] == 0 {
                queue = append(queue, []int{i, j})
            }
        }
    }
    ans := 0
    for len(queue) != 0 {
        ans++
        size := len(queue)
        for i := 0; i < size; i++ {
            cell := queue[0]
            queue = queue[1:]
            row, column := cell[0], cell[1]
            for _, dir := range dirs {
                newRow, newColumn := row + dir[0], column + dir[1]
                if newRow >= 0 && newRow < rows && newColumn >= 0 && newColumn < columns && matrix[newRow][newColumn] < matrix[row][column] {
                    outdegrees[newRow][newColumn]--
                    if outdegrees[newRow][newColumn] == 0 {
                        queue = append(queue, []int{newRow, newColumn})
                    }
                }
            }
        }
    }
    return ans
}

golang 解法, 执行用时: 24 ms, 内存消耗: 6.7 MB, 提交时间: 2023-04-22 12:27:53

var (
    dirs = [][]int{[]int{-1, 0}, []int{1, 0}, []int{0, -1}, []int{0, 1}}
    rows, columns int
)

func longestIncreasingPath(matrix [][]int) int {
    if len(matrix) == 0 || len(matrix[0]) == 0 {
        return 0
    }
    rows, columns = len(matrix), len(matrix[0])
    memo := make([][]int, rows)
    for i := 0; i < rows; i++ {
        memo[i] = make([]int, columns)
    }
    ans := 0
    for i := 0; i < rows; i++ {
        for j := 0; j < columns; j++ {
            ans = max(ans, dfs(matrix, i, j, memo))
        }
    }
    return ans
}

func dfs(matrix [][]int, row, column int, memo [][]int) int {
    if memo[row][column] != 0 {
        return memo[row][column]
    }
    memo[row][column]++
    for _, dir := range dirs {
        newRow, newColumn := row + dir[0], column + dir[1]
        if newRow >= 0 && newRow < rows && newColumn >= 0 && newColumn < columns && matrix[newRow][newColumn] > matrix[row][column] {
            memo[row][column] = max(memo[row][column], dfs(matrix, newRow, newColumn, memo) + 1)
        }
    }
    return memo[row][column]
}

func max(x, y int) int {
    if x > y {
        return x
    }
    return y
}

python3 解法, 执行用时: 272 ms, 内存消耗: 19.1 MB, 提交时间: 2023-04-22 12:27:40

class Solution:
    
    DIRS = [(-1, 0), (1, 0), (0, -1), (0, 1)]

    def longestIncreasingPath(self, matrix: List[List[int]]) -> int:
        if not matrix:
            return 0
        
        @lru_cache(None)
        def dfs(row: int, column: int) -> int:
            best = 1
            for dx, dy in Solution.DIRS:
                newRow, newColumn = row + dx, column + dy
                if 0 <= newRow < rows and 0 <= newColumn < columns and matrix[newRow][newColumn] > matrix[row][column]:
                    best = max(best, dfs(newRow, newColumn) + 1)
            return best

        ans = 0
        rows, columns = len(matrix), len(matrix[0])
        for i in range(rows):
            for j in range(columns):
                ans = max(ans, dfs(i, j))
        return ans

java 解法, 执行用时: 8 ms, 内存消耗: 41.6 MB, 提交时间: 2023-04-22 12:27:05

// 记忆化深度优先搜索
class Solution {
    public int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
    public int rows, columns;

    public int longestIncreasingPath(int[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return 0;
        }
        rows = matrix.length;
        columns = matrix[0].length;
        int[][] memo = new int[rows][columns];
        int ans = 0;
        for (int i = 0; i < rows; ++i) {
            for (int j = 0; j < columns; ++j) {
                ans = Math.max(ans, dfs(matrix, i, j, memo));
            }
        }
        return ans;
    }

    public int dfs(int[][] matrix, int row, int column, int[][] memo) {
        if (memo[row][column] != 0) {
            return memo[row][column];
        }
        ++memo[row][column];
        for (int[] dir : dirs) {
            int newRow = row + dir[0], newColumn = column + dir[1];
            if (newRow >= 0 && newRow < rows && newColumn >= 0 && newColumn < columns && matrix[newRow][newColumn] > matrix[row][column]) {
                memo[row][column] = Math.max(memo[row][column], dfs(matrix, newRow, newColumn, memo) + 1);
            }
        }
        return memo[row][column];
    }
}

上一题

下一题

详情