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English Version

题目描述

给你一个二维字符网格数组 grid ,大小为 m x n ,你需要检查 grid 中是否存在 相同值 形成的环。

一个环是一条开始和结束于同一个格子的长度 大于等于 4 的路径。对于一个给定的格子,你可以移动到它上、下、左、右四个方向相邻的格子之一,可以移动的前提是这两个格子有 相同的值 

同时,你也不能回到上一次移动时所在的格子。比方说,环  (1, 1) -> (1, 2) -> (1, 1) 是不合法的,因为从 (1, 2) 移动到 (1, 1) 回到了上一次移动时的格子。

如果 grid 中有相同值形成的环,请你返回 true ,否则返回 false 。

 

示例 1:

输入:grid = [["a","a","a","a"],["a","b","b","a"],["a","b","b","a"],["a","a","a","a"]]
输出:true
解释:如下图所示,有 2 个用不同颜色标出来的环:

示例 2:

输入:grid = [["c","c","c","a"],["c","d","c","c"],["c","c","e","c"],["f","c","c","c"]]
输出:true
解释:如下图所示,只有高亮所示的一个合法环:

示例 3:

输入:grid = [["a","b","b"],["b","z","b"],["b","b","a"]]
输出:false

 

提示:

  • m == grid.length
  • n == grid[i].length
  • 1 <= m <= 500
  • 1 <= n <= 500
  • grid 只包含小写英文字母。

解法

构造并查集,遍历每个坐标 (i, j),如果下方或者右侧的元素 (x, y) 与当前元素 (i, j) 相同,进行合并操作。若是,若此前两个坐标已经处于连通状态,再进行合并时会形成环,直接返回 true。否则遍历结束返回 false。

并查集模板:

模板 1——朴素并查集:

# 初始化,p存储每个点的父节点
p = list(range(n))

# 返回x的祖宗节点
def find(x):
    if p[x] != x:
        # 路径压缩
        p[x] = find(p[x])
    return p[x]


# 合并a和b所在的两个集合
p[find(a)] = find(b)

模板 2——维护 size 的并查集:

# 初始化,p存储每个点的父节点,size只有当节点是祖宗节点时才有意义,表示祖宗节点所在集合中,点的数量
p = list(range(n))
size = [1] * n

# 返回x的祖宗节点
def find(x):
    if p[x] != x:
        # 路径压缩
        p[x] = find(p[x])
    return p[x]

# 合并a和b所在的两个集合
if find(a) != find(b):
    size[find(b)] += size[find(a)]
    p[find(a)] = find(b)

模板 3——维护到祖宗节点距离的并查集:

# 初始化,p存储每个点的父节点,d[x]存储x到p[x]的距离
p = list(range(n))
d = [0] * n

# 返回x的祖宗节点
def find(x):
    if p[x] != x:
        t = find(p[x])
        d[x] += d[p[x]]
        p[x] = t
    return p[x]

# 合并a和b所在的两个集合
p[find(a)] = find(b)
d[find(a)] = distance

Python3

class Solution:
    def containsCycle(self, grid: List[List[str]]) -> bool:
        def find(x):
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        m, n = len(grid), len(grid[0])
        p = list(range(m * n))
        for i in range(m):
            for j in range(n):
                for a, b in [[0, 1], [1, 0]]:
                    x, y = i + a, j + b
                    if x < m and y < n and grid[x][y] == grid[i][j]:
                        if find(x * n + y) == find(i * n + j):
                            return True
                        p[find(x * n + y)] = find(i * n + j)
        return False

Java

class Solution {
    private int[] p;

    public boolean containsCycle(char[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        p = new int[m * n];
        for (int i = 0; i < p.length; ++i) {
            p[i] = i;
        }
        int[] dirs = {0, 1, 0};
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                for (int k = 0; k < 2; ++k) {
                    int x = i + dirs[k];
                    int y = j + dirs[k + 1];
                    if (x < m && y < n && grid[i][j] == grid[x][y]) {
                        if (find(x * n + y) == find(i * n + j)) {
                            return true;
                        }
                        p[find(x * n + y)] = find(i * n + j);
                    }
                }
            }
        }
        return false;
    }

    private int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
}

C++

class Solution {
public:
    vector<int> p;

    bool containsCycle(vector<vector<char>>& grid) {
        int m = grid.size(), n = grid[0].size();
        p.resize(m * n);
        for (int i = 0; i < p.size(); ++i) p[i] = i;
        vector<int> dirs = {0, 1, 0};
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                for (int k = 0; k < 2; ++k) {
                    int x = i + dirs[k], y = j + dirs[k + 1];
                    if (x < m && y < n && grid[x][y] == grid[i][j]) {
                        if (find(x * n + y) == find(i * n + j)) return 1;
                        p[find(x * n + y)] = find(i * n + j);
                    }
                }
            }
        }
        return 0;
    }

    int find(int x) {
        if (p[x] != x) p[x] = find(p[x]);
        return p[x];
    }
};

Go

func containsCycle(grid [][]byte) bool {
	m, n := len(grid), len(grid[0])
	p := make([]int, m*n)
	for i := range p {
		p[i] = i
	}
	var find func(x int) int
	find = func(x int) int {
		if p[x] != x {
			p[x] = find(p[x])
		}
		return p[x]
	}
	dirs := []int{1, 0, 1}
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			for k := 0; k < 2; k++ {
				x, y := i+dirs[k], j+dirs[k+1]
				if x < m && y < n && grid[x][y] == grid[i][j] {
					if find(x*n+y) == find(i*n+j) {
						return true
					}
					p[find(x*n+y)] = find(i*n + j)
				}
			}
		}
	}
	return false
}

JavaScript

/**
 * @param {character[][]} grid
 * @return {boolean}
 */
var containsCycle = function (grid) {
    const m = grid.length;
    const n = grid[0].length;
    let p = Array.from({ length: m * n }, (_, i) => i);
    function find(x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
    const dirs = [0, 1, 0];
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            for (let k = 0; k < 2; ++k) {
                const x = i + dirs[k];
                const y = j + dirs[k + 1];
                if (x < m && y < n && grid[x][y] == grid[i][j]) {
                    if (find(x * n + y) == find(i * n + j)) {
                        return true;
                    }
                    p[find(x * n + y)] = find(i * n + j);
                }
            }
        }
    }
    return false;
};

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