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2045:到达目的地的第二短时间（添加代码注释）

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轩辕龙儿 2022-01-24 15:23:53 +08:00
parent a09e96a85c
commit 6d397ff0d9
1 changed files with 19 additions and 15 deletions
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32
src/main/java/leetcode/editor/cn/SecondMinimumTimeToReachDestination.java
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@ -92,6 +92,7 @@ public class SecondMinimumTimeToReachDestination {
//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
public int secondMinimum(int n, int[][] edges, int time, int change) {
// 统计所有节点的联通节点，并将其存入map中留着后面使用
Map<Integer, List<Integer>> map = new HashMap<>(n);
for (int i = 1; i <= n; i++) {
map.put(i, new ArrayList<>());
@ -100,36 +101,39 @@ public class SecondMinimumTimeToReachDestination {
map.get(edge[0]).add(edge[1]);
map.get(edge[1]).add(edge[0]);
}
Queue<int[]> queue = new LinkedList<>();
queue.add(new int[]{1, 0});
Queue<Integer> queue = new LinkedList<>();
queue.add(1);
// 记录节点到达的次数
int[] counts = new int[n + 1];
// 记录到达节点的时间
int free = 0;
while (!queue.isEmpty()) {
int[] arrs = queue.peek();
int free = arrs[1] + time;
if ((arrs[1] / change) % 2 == 1) {
free += change - arrs[1] % change;
// 红灯情况下加上需要等待的时间
if (free % (2 * change) >= change) {
free += change - free % change;
}
free += time;
// 同一时间可以到达的节点数量
int size = queue.size();
boolean bl = true;
// 同一时间节点是否已经到达
boolean[] use = new boolean[n + 1];
for (int i = 0; i < size; i++) {
arrs = queue.poll();
List<Integer> list = map.get(arrs[0]);
// 获取该节点接下来可以到达的节点
List<Integer> list = map.get(queue.poll());
for (int num : list) {
// 同一时间未到达，并且到达该节点的总次数小于2
if (!use[num] && counts[num] < 2) {
queue.add(new int[]{num, free});
queue.add(num);
use[num] = true;
counts[num]++;
}
if (num == n && bl) {
bl = false;
if (counts[num] == 2) {
// 如果是第二次到达最后一个节点，直接返回需要到达的诗句
if (num == n && counts[num] == 2) {
return free;
}
}
}
}
}
return 0;
}
}