1129. 颜色交替的最短路径
问题1129 颜色交替的最短路径 ★★☆☆☆
https://leetcode.cn/problems/shortest-path-with-alternating-colors/
这个题目可是费了不少的时间和精力。从解题入手点来说，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。
最终只得参考了下官方题解，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，但是代码看起来有点复杂，不过相比官方题解的代码，我又觉得不如我自己写的好理解。
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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()) {
int v = queue.poll();
if (shortestPathLengthWithBlueLastPath[v] != -1) {
int length = shortestPathLengthWithBlueLastPath[v] + 1;
for (int[] redEdge : redEdges) {
if (redEdge[0] == v) {
int next = redEdge[1];
if (shortestPathLengthWithRedLastPath[next] == -1) {
shortestPathLengthWithRedLastPath[next] = length;
queue.add(next);
} else {
if (length < shortestPathLengthWithRedLastPath[next]) {
shortestPathLengthWithRedLastPath[next] = length;
}
}
}
}
}
if (shortestPathLengthWithRedLastPath[v] != -1) {
int length = shortestPathLengthWithRedLastPath[v] + 1;
for (int[] blueEdge : blueEdges) {
if (blueEdge[0] == v) {
int next = blueEdge[1];
if (shortestPathLengthWithBlueLastPath[next] == -1) {
shortestPathLengthWithBlueLastPath[next] = length;
queue.add(next);
} else {
if (length < shortestPathLengthWithBlueLastPath[next]) {
shortestPathLengthWithBlueLastPath[next] = length;
}
}
}
}
}
}
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;
}