题目链接

https://leetcode.cn/problems/shortest-path-with-alternating-colors/

解答过程

这个题目可是费了不少的时间和精力。从解题着手点来说，大的思路并不难，比较容易想到可以通过广度优先遍历（BFS）来解决这个问题。问题的关键在于每一层BFS具体做哪些操作？这可以从循环不变量来入手。我一开始的思路是每次进入循环时，当前维护着这样两个变量：shortestPathLength[]，记录的是从0节点到v节点的最短颜色交错路径长度；lastPathColor[]，记录的是从0节点到v节点的最短颜色交错路径的最后一条path的颜色，可能的取值有不可达、红色、蓝色、红蓝均可等四种情况。在当前遍历中，根据当前v节点的lastPathColor，决定是继续遍历redEdges或者blueEdges或者都遍历，同时更新遍历到的节点的shortestPath和lastPathColor。直到遍历结束，根据lastPathColor确定不可达的节点并更新shortestPathLength，然后返回。整个思路看起来还挺正确吧？处理了一些细节问题以后，最终的结果是超时，出现了死循环，因为可能出现环，对于已经遍历过的节点是否要二次遍历直觉上很难处理。

这个时候只能参考官方题解了。看了一下官方题解，大的思路是类似的，但有一个地方启发了我，那就是分别维护shortestPathLengthWithRedLastPath[]和shortestPathLengthWithBlueLastPath[]，而不是妄图用一个lastPathColor[]来维护所有可能的状态。修改了一下代码，终于通过了，但是代码看起来属实有点复杂。

	public int[] shortestAlternatingPaths(int n, int[][] redEdges, int[][] blueEdges) {
		int[] shortestPathLengthWithRedLastPath = new int[n];
		shortestPathLengthWithRedLastPath[0] = 0;
		for (int i = 1; i < n; i++) {
			shortestPathLengthWithRedLastPath[i] = -1;
		}

		int[] shortestPathLengthWithBlueLastPath = new int[n];
		shortestPathLengthWithBlueLastPath[0] = 0;
		for (int i = 1; i < n; i++) {
			shortestPathLengthWithBlueLastPath[i] = -1;
		}

		Queue<Integer> queue = new LinkedList<>();
		queue.offer(0);

		while (!queue.isEmpty()) {
			Set<Integer> nextRound = new HashSet<>();

			int size = queue.size();
			while (size-- > 0) {
				int v = queue.poll();

				if (shortestPathLengthWithBlueLastPath[v] != -1) {
					int length = shortestPathLengthWithBlueLastPath[v] + 1;
					for (int i = 0; i < redEdges.length; i++) {
						if (redEdges[i][0] == v) {
							int next = redEdges[i][1];

							if (shortestPathLengthWithRedLastPath[next] == -1) {
								shortestPathLengthWithRedLastPath[next] = length;
								nextRound.add(next);
							} else {
								if (length < shortestPathLengthWithRedLastPath[next]) {
									shortestPathLengthWithRedLastPath[next] = length;
								}
							}
						}
					}
				}

				if (shortestPathLengthWithRedLastPath[v] != -1) {
					int length = shortestPathLengthWithRedLastPath[v] + 1;
					for (int i = 0; i < blueEdges.length; i++) {
						if (blueEdges[i][0] == v) {
							int next = blueEdges[i][1];

							if (shortestPathLengthWithBlueLastPath[next] == -1) {
								shortestPathLengthWithBlueLastPath[next] = length;
								nextRound.add(next);
							} else {
								if (length < shortestPathLengthWithBlueLastPath[next]) {
									shortestPathLengthWithBlueLastPath[next] = length;
								}
							}
						}
					}
				}
			}

			queue.addAll(nextRound);
		}

		int[] shortestPathLength = new int[n];
		for (int i = 0; i < n; i++) {
			int withRedLastPath = shortestPathLengthWithRedLastPath[i];
			int withBlueLastPath = shortestPathLengthWithBlueLastPath[i];

			if (withRedLastPath == -1
				&& withBlueLastPath == -1) {
				shortestPathLength[i] = -1;
			} else if (withRedLastPath == -1) {
				shortestPathLength[i] = withBlueLastPath;
			} else if (withBlueLastPath == -1) {
				shortestPathLength[i] = withRedLastPath;
			} else {
				shortestPathLength[i] = Math.min(withRedLastPath, withBlueLastPath);
			}
		}

		return shortestPathLength;
	}

又看了一些官方题解的代码，额，看不懂，虽然是看着简洁一些，但感觉还不如我自己写的好理解。